Principles of Database Systems

Transcript

1 Principles of Database Systems V. Megalooikonomou Συναρτησιακές Εξαρτήσεις (Functional Dependencies) (based on notes by Silberchatz,Korth, and Sudarshan and notes by C. Faloutsos)

2 General Overview Formal query languages rel algebra and calculi Commercial query languages SQL QBE, (QUEL) Integrity constraints Συναρτησιακές Εξαρτήσεις Normalization - good DB design

3 Overview Domain; Ref. Integrity constraints Assertions and Triggers Security Συναρτησιακές Εξαρτήσεις why definition Armstrong s axioms closure and cover

4 Συναρτησιακές Εξαρτήσεις motivation: good tables takes1 (ssn, c-id, grade, name, address) good or bad?

5 Συναρτησιακές Εξαρτήσεις takes1 (ssn, c-id, grade, name, address) Ssn c -id G rade Name A ddress A s mith Main B s mith Main A s mith Main

6 Συναρτησιακές Εξαρτήσεις Bad - why? Ssn c -id G rade Name A ddress A s mith Main B s mith Main A s mith Main

7 Συναρτησιακές Εξαρτήσεις Redundancy space inconsistencies insertion/deletion anomalies (later ) What caused the problem?

8 Συναρτησιακές Εξαρτήσεις name depends on ssn define depends Ssn c -id G rade Name A ddress A s mith Main B s mith Main A s mith Main

9 Συναρτησιακές Εξαρτήσεις a b Definition: a functionally determines b ( a συναρτησιακά καθορίζει το b ) Ssn c -id G rade Name A ddress A s mith Main B s mith Main A s mith Main

10 Συναρτησιακές Εξαρτήσεις Informally: if you know a, there is only one b to match Ssn c -id G rade Name A ddress A s mith Main B s mith Main A s mith Main

11 Συναρτησιακές Εξαρτήσεις formally: Y ( t1[ x] = t2[ x] t1[ y] = t2[ y]) if two tuples agree on the attribute, they *must* agree on the Y attribute, too (e.g., if ssn is the same, so should address) a functional dependency is a generalization of the notion of a key -Why?

12 Συναρτησιακές Εξαρτήσεις, Y can be sets of attributes other examples?? Ssn c -id G rade Name A ddress A s mith Main B s mith Main A s mith Main

13 Συναρτησιακές Εξαρτήσεις ssn -> name, address ssn, c-id -> grade Ssn c -id G rade Name A ddress A s mith Main B s mith Main A s mith Main

14 Συναρτησιακές Εξαρτήσεις K is a superkey for relation R iff K -> R K is a candidate key for relation R iff: K -> R for no a K, a -> R

15 Συναρτησιακές Εξαρτήσεις Closure (κλειστότητα) of a set of FD: all implied FDs e.g.: ssn -> name, address ssn, c-id -> grade imply (συνάγουν) ssn, c-id -> grade, name, address ssn, c-id -> ssn

16 FDs - Armstrong s axioms Closure of a set of FD: all implied FDs e.g.: ssn -> name, address ssn, c-id -> grade how to find all the implied ones, systematically?

17 FDs - Armstrong s axioms Armstrong s axioms guarantee soundness and completeness: Reflexivity Y Y (ανακλαστικότητα): e.g., ssn, name -> ssn Augmentation Y W YW (επαυξητικότητα): e.g., ssn->name then ssn,grade-> ssn,grade

18 FDs - Armstrong s axioms Transitivity (μεταβατικότητα) Y Y Z Z ssn->address address-> county-tax-rate THEN: ssn-> county-tax-rate

19 FDs - Armstrong s axioms Reflexivity: Y Y Augmentation: Y W YW Transitivity: Y Y Z Z sound and complete

20 FDs finding the closure F+ F + = F repeat for each functional dependency f in F + apply reflexivity and augmentation rules on f add the resulting Συναρτησιακές Εξαρτήσεις to F + for each pair of Συναρτησιακές Εξαρτήσεις f 1 and f 2 in F + if f 1 and f 2 can be combined using transitivity then add the resulting functional dependency to F + until F + does not change any further We can further simplify manual computation of F + by using the following additional rules

21 FDs - Armstrong s axioms Additional rules: Union (ένωση) Decomposition (διασπαστικότητα) Pseudo-transitivity (ψευδομεταβατικότητα) Y Z YZ YW Y YZ Z Y W Z Z

22 FDs - Armstrong s axioms Prove Union from the three axioms: Y? Z YZ

23 FDs - Armstrong s axioms Prove Union from the three axioms: Y (1) Z (2) (1) + augm. w/ Z (2) + augm. w/ but (3) + (4) is ; thus and Z transitivity YZ Z (3) (4) YZ

24 FDs - Armstrong s axioms Prove Pseudo-transitivity: Y Y W Y YW YW? Y Z W Z Y Y Z Z

25 FDs - Armstrong s axioms Prove Decomposition Y Y W Y YW YZ? Y Z Y Y Z Z

26 FDs - Closure F+ Given a set F of FD (on a schema) F+ is the set of all implied FD. E.g., takes(ssn, c-id, grade, name, address) ssn, c-id -> grade ssn-> name, address }F

27 FDs - Closure F+ ssn, c-id -> grade ssn-> name, address ssn-> ssn ssn, c-id-> address c-id, address-> c-id... F+

28 FDs - Closure F+ R=(A,B,C,G,H,I) F= { A->B A->C CG->H CG->I B->H} Some members of F+: A->H AG->I CG->HI

29 FDs - Closure A+ Given a set F of FD (on a schema) A+ is the set of all attributes determined by A: takes(ssn, c-id, grade, name, address) ssn, c-id -> grade ssn-> name, address }F {ssn}+ =??

30 FDs - Closure A+ takes(ssn, c-id, grade, name, address) ssn, c-id -> grade ssn-> name, address }F {ssn}+ ={ssn, name, address }

31 FDs - Closure A+ takes(ssn, c-id, grade, name, address) ssn, c-id -> grade ssn-> name, address }F {c-id}+ =??

32 FDs - Closure A+ takes(ssn, c-id, grade, name, address) ssn, c-id -> grade ssn-> name, address }F {c-id, ssn}+ =??

33 FDs - Closure A+ if A+ = {all attributes of table} then A is a candidate key

34 FDs - Closure A+ Algorithm to compute α +, the closure of α under F result := α; while (changes to result) do for each β γin F do begin if β resul then result := result γ end

35 FDs - Closure A+ (example) R = (A, B, C, G, H, I) F = {A B, A C, CG H, CG I, B H} (AG) + 1. result = AG 2. result = ABCG (A C and A B) 3. result = ABCGH(CG H and CG AGBC) 4. result = ABCGHI (CG I and CG AGBCH) Is AG a candidate key? 1. Is AG a super key? 1. Does AG R? 2. Is any subset of AG a superkey? 1. Does A + R? 2. Does G + R?

36 FDs - A+ closure Diagrams AB->C (1) A->BC (2) B->C (3) A->B (4) A B C

37 FDs - canonical cover Fc Given a set F of FD (on a schema) Fc (ελάχιστο κάλλυμα) is a minimal set of equivalent FD. E.g., takes(ssn, c-id, grade, name, address) ssn, c-id -> grade ssn-> name, address ssn,name-> name, address ssn, c-id-> grade, name F

38 FDs - canonical cover Fc Fc ssn, c-id -> grade ssn-> name, address ssn,name-> name, address ssn, c-id-> grade, name F

39 FDs - canonical cover Fc why do we need it? define it properly compute it efficiently

40 FDs - canonical cover Fc why do we need it? easier to compute candidate keys define it properly compute it efficiently

41 FDs - canonical cover Fc define it properly - three properties every FD a->b has no extraneous attributes on the RHS same for the LHS all LHS parts are unique

42 FDs - canonical cover Fc extraneous attribute: if the closure is the same, before and after its elimination or if F-before implies F-after and vice-versa

43 FDs - canonical cover Fc ssn, c-id -> grade ssn-> name, address ssn,name-> name, address ssn, c-id-> grade, name F

44 FDs - canonical cover Fc Algorithm: examine each FD; drop extraneous LHS or RHS attributes merge FDs with same LHS repeat until no change

45 FDs - canonical cover Fc Trace algo for AB->C (1) A->BC (2) B->C (3) A->B (4)

46 FDs - canonical cover Fc Trace algo for AB->C (1) A->BC (2) B->C (3) A->B (4) (4) and (2) merge: AB->C (1) A->BC (2) B->C (3)

47 FDs - canonical cover Fc AB->C (1) A->BC (2) B->C (3) AB->C (1) A->B (2 ) B->C (3) in (2): C is extr.

48 FDs - canonical cover Fc AB->C (1) A->B (2 ) B->C (3) B->C (1 ) A->B (2 ) B->C (3) in (1): A is extr.

49 FDs - canonical cover Fc B->C (1 ) A->B (2 ) B->C (3) (1 ) and (3) merge A->B (2 ) B->C (3) nothing is extraneous: canonical cover

50 FDs - canonical cover Fc BEFORE AB->C (1) A->BC (2) B->C (3) A->B (4) AFTER A->B (2 ) B->C (3)

51 Overview - conclusions Domain; Ref. Integrity constraints Assertions and Triggers Συναρτησιακές Εξαρτήσεις why definition Armstrong s axioms closure and cover