Lecture 15 10/02/15. CMPSC431W: Database Management Systems. Instructor: Yu- San Lin

Transcription

1 CMPSC431W: Database Management Systems Lecture 15 10/02/15 Instructor: Yu- San Lin Course Website: hcp:// Slides based on McGraw- Hill & Dr. Wang- Chien Lee 1

2 Review Reflexivity If Y X, then X à Y AugmentaXon If X à Y, then XZ à YZ for any Z TransiXvity If X à Y and Y à Z, then X à Z Union If X à Y and X à Z, then X à YZ DecomposiXon If X à YZ, then X à Y and X à Z Armstrong s Axioms 2

3 Example from last Xme Contracts (contracxd, supplierid, projecxd, depxd, parxd, qty, value) Schema: CSJDPQV Given ICs: 1. C is the key: C à CSJDPQV 2. Project purchases a part using single contract: JP à C 3. Department purchases at most one part from a supplier: SD à P 3

4 Example for applying Armstrong s Axioms (cont.) JP à C, C à CSJDPQV (transixvity) JP à CSJDPQV So that one can say JP is also a key SD à P (augmentaxon) SDJ à JP SDJ à JP, JP à CSJDPQV (transixvity) SDJ à CSJDPQV Can we do this? SDJ à CSJDPQV No! 4

5 Theorem 1 Armstrong s Axioms are: Sound: they generate only FDs in F + when applied to a set F of FDs Complete: repeated applicaxon of these rules will generate all FDs in the closure F + i.e., F f iff F f * The soundness is straigh2orward to prove; completeness is harder to show. 5

6 Prove the Soundness of Union 6

7 CompuXng the closure of FDs CompuXng the F + of F can be expensive. Size of closure is Typically, we just want to check if a given FD X à Y is in the closure of a set of FDs F An efficient check: Compute acribute closure of X (denoted X + ) w.r.t. F. Set of all acributes A such that X à A is in F + Check if Y is in X + 7

8 ACribute Closure Algorithm for compuxng the acribute closure: closure = X; repeat until there is no change: { if there is an FD U! V in F such that U closure, then set closure = closure V } 8

9 Example of ACribute Closure Example: Does F = {A à B, B à C, CD à E} imply A à E? i.e., is A à E in the closure F +? Equivalently, is E in A + w.r.t. F? 9

10 Normal Forms Normal forms help to find problems that might arise from the current schema If a relaxon is in a certain normal form, certain kinds of problems arise First normal form (1NF) Second normal form (2NF) Third normal form (3NF) Boyce- Codd normal form (BCNF) * E.g., every relaxon in BCNF is also in 3NF. 10

11 First Normal Form (1NF) A relaxon is in first normal form if every field contains only (no lists or sets) è An implicit requirement in the relaxonal model Which one below is in 1NF? dno dname loca1ons 4 Sales {Boston, New York, Chicago} 1 Research {Boston} dno dname loca1ons 4 Sales Boston 4 Sales New York 4 Sales Chicago 1 Research Boston 11

12 Second Normal Form (2NF) X à Y is: funcxonal dependency: if removing acribute A from X makes X à Y doesn t hold funcxonal dependency: otherwise A relaxon R is in 2NF if every non- key acribute in R is fully funcxonally dependent on the candidate key of R A non- 2NF schema can be decomposed to a 2NF 12

13 Decompose into 2NF Which of the following acribute are not fully FD on primary key? Emp_Proj ssn pno hours ename plocaxon How to decompose? 13

14 Second Normal Form (2NF) (cont.) IS this in 2NF? ename ssn bdate addr dno dname dmgr There could sxll exist a non- key acribute funcxonally dependent on a non- key acribute 14

15 Third Normal Form (3NF) RelaXon R with FDs F is in 3NF if, for all X à A in F+: 1) A X (trivial FD), OR 2) X contains a key for R, i.e., X is a superkey, OR 3) A is part of some key for R 2) 3) X key key A 15

16 Third Normal Form (3NF) (cont.) R should not have a non- key acribute funcxonally determined by another non- key acribute If R is in 3NF, some redundancy is sxll possible Use when BCNF is not achievable 16

17 Third Normal Form (3NF) (cont.) If 3NF violated by (2) for an X à A, X contains a key, one of the following holds: ParXal dependency: X is a subset of some key K (X, A) pairs stored redundantly SoluXon: TransiXve dependency: X is not a proper subset of any key Exists chain K à X à A, (X, A) stored redundantly E.g., (ssn à raxng à hourly_wage) 17

18 Third Normal Form (3NF) (cont.) ParXal dependency key X A Case 1: A not in KEY TransiXve dependency key X A Case 1: A not in KEY key A X Case 2: A is in KEY 18

19 Third Normal Form (3NF) (cont.) Even if a relaxon is in 3NF, some redundancy is sxll possible Example: sid bid date card 19

20 Decompose into 3NF ename ssn bdate addr dno dname dmgr 20