CS 319: Theory of Databases: FDs

1. CS 319: Theory of Databases: FDs Dr. A.I. Cristea http://www.dcs.warwick.ac.uk/~acristea/

2. … previous Generalities on Databases

3. … previous Generalities on Databases Definitions of databases The issues databases tried/try to solve The ingredients of a database The users of a database and their respective roles (look at the later review of the database administrator as well) The data abstraction levels in a database The data models in a database The distinction between instance and schema Data definition versus data manipulation language Data manager program and its functions Overall database system structure

4. Content • Generalities DB • Integrity constraints (FD revisited) • Relational Algebra (revisited) • Query optimisation • Tuple calculus • Domain calculus • Query equivalence • LLJ, DP and applications • Temporal Data • The Askew Wall

5. Functional Dependency • A functional dependency (FD) has the form X  Ywhere X and Y are sets of attributes in a relation R X  Y iff any two tuples that agree on X value also agree on Y value X  Y if and only if: for any instance r of R for any tuples t1 and t2 of r t1(X) = t2(X)  t1(Y) = t2(Y) also written: (r  R, t1, t2  r : t1[X]=t2[X] :t1[Y]=t2[Y]) basically identical with: (r  R,  t1, t2  r ::t1[X]=t2[X]  t1[Y]=t2[Y])

6. Notations • r R indicates instance r is a valid instance for schema R (relation type). • t  r indicates t is a tuple of r. • X  R(***) indicates X is a subset of the set of attributes used by R (~ heading). • XYmeans X  Y. (***) Should actually be X  Attr(R) (heading)

7. To prove or not to prove, that is the question. Given a proposition Q it always holds that Q  Q. For example: {De Morgan} ergo Prove or give a counter example

8. Proving • To prove a functional dependency we can use the inference rules (Armstrong) or the definition of functional dependency • Normally, the choice is optional.

9. Why prove something using definition of FD? • Usually we prefer “inference rules”. • However: we must prove that they are correct (hold). • via FD definitions!

10. Ex: Augmentation and Transitivity rules • Augmentation: Prove (using the definition of fd) that if X, Y and Z are sets of attributes of a relational schema R, and the fd X  Y holds in R, then XZ  YZ also holds in R. • Transitivity: Prove (using the definition of fd) that if X, Y and Z are sets of attributes of a relational schema R, and the fds X  Y and Y  Z hold in R, then X  Z also holds in R.

11. Augumentation (in short) (r  R, t1, t2  r : t1[X]=t2[X] : t1[Y]=t2[Y]) (this is the definition of X  Y) (r  R,  t1, t2  r : t1[Z]=t2[Z] : t1[Z]=t2[Z]) (this is always true) (because ((A  B) Ù (C  D)  (A Ù C)  (B Ù D)) (r  R,  t1, t2  r : t1[X]=t2[X] Ù t1[Z]=t2[Z] : t1[Y]=t2[Y] Ù t1[Z]=t2[Z]) (because for a function t: t[X  Z] = t[X]  t[Z]) (r  R,  t1, t2  r : t1[XZ]=t2[XZ] : t1[YZ]=t2[YZ]) (this is the definition of XZ  YZ)

12. Lemma 1 • (((A  B) Ù (C  D)) (A Ù C)  (B Ù D)) • <=> (use (X => Y) <=> (~X v Y) (twice) and distribute the negation over the conjunction) ~(A  B) v ~(C  D) v ~(A Ù C) v (B Ù D) • <=> (use ~(X => Y) <=> (X Ù ~Y), distribute negation over conjunction) (A Ù ~B) v (C Ù ~D) v ~A v ~C v (B Ù D) • <=> (use ((X Ù ~Y) v ~X) <=> (~Y v ~X)) ~A v~B v~C v~D v (B Ù D) • <=> (distribute negation over conjunction) ~A v~(B Ù D) v~C v (B Ù D) • <=> ( (X v~X) <=> true; true/false elimination) true

13. Augumentation (formal -1) • (r  R,  t1, t2  r ::t1[X]=t2[X]  t1[Y]=t2[Y]) (this is the definition of X  Y) • (r  R,  t1, t2  r :: t1[Z]=t2[Z]  t1[Z]=t2[Z]) (this is always true) • Since both (1) and (2) hold, we can conjugate them: (r  R,  t1, t2  r :: t1[X]=t2[X]  t1[Y]=t2[Y]) Ù (r  R,  t1, t2  r :: t1[Z]=t2[Z]  t1[Z]=t2[Z]) • (domain splitting) • (r  R,  t1, t2  r :: (t1[X]=t2[X]  t1[Y]=t2[Y]) Ù (t1[Z]=t2[Z]  t1[Z]=t2[Z]))

14. Augumentation (formal -2) • (domain splitting) • (r  R,  t1, t2  r :: (t1[X]=t2[X]  t1[Y]=t2[Y]) Ù (t1[Z]=t2[Z]  t1[Z]=t2[Z])) •  (because of Lemma 1: ((A  B) Ù (C  D))  ((A Ù C)  (B Ù D))) (r  R,  t1, t2  r :: (t1[X]=t2[X] Ù t1[Z]=t2[Z])  (t1[Y]=t2[Y] Ù t1[Z]=t2[Z])) •  (because for a function t: t[X  Z] = t[X]  t[Z]) (r  R,  t1, t2  r :: (t1[XZ]=t2[XZ]  t1[YZ]=t2[YZ]) (this is the definition of XZ  YZ)

15. Transitivity (1) • (1) (r  R,  t1, t2  r:: (t1[X]=t2[X])  (t1[Y]=t2[Y])) (definition of X  Y) • (2) (r  R,  t1, t2  r:: (t1[Y]=t2[Y])  (t1[Z]=t2[Z])) (definition of Y  Z) • Since both (1) and (2) hold, we can conjugate them: (r  R,  t1, t2  r :: (t1[X]=t2[X])  (t1[Y]=t2[Y])) Ù (r  R,  t1, t2  r :: (t1[Y]=t2[Y])  (t1[Z]=t2[Z])) • (domain splitting) (r  R,  t1, t2  r ::( t1[X]=t2[X]  t1[Y]=t2[Y]) Ù (t1[Y]=t2[Y]  t1[Z]=t2[Z]))

16. Transitivity (2) • (domain splitting) (r  R,  t1, t2  r ::( t1[X]=t2[X]  t1[Y]=t2[Y]) Ù (t1[Y]=t2[Y]  t1[Z]=t2[Z])) • (because of Lemma 2: ((A  B) Ù (B  C)) (A  C)) • (r  R,  t1, t2  r :: t1[X]=t2[X]  t1[Z]=t2[Z]) (this is the definition of X  Z)

17. Lemma 2 • ((A  B) Ù (B  C)) (A  C) • (use (X  Y)  (~X v Y) and distribute the negation over the conjunction) ~(A v B) v ~(B v C) v (~A v C) • (use ~(X  Y)  (X Ù ~Y), distribute negation over conjunction) (A Ù ~B) v (B Ù ~C) v (~A v C) • (use ((X Ù ~Y) v ~X)  (~Y v ~X)) ~A v ~B v B v C • ( (X v ~X)  true; true/false elimination) true

18. Disproving • to show a rule does not hold you must find (using your imagination) at least one instance in which the given fds hold and the “supposedly implied” fds do not hold.

19. Bogus rules • Disprove that if X and Y are sets of attributes of a relational schema R, and the fd X  Y holds in R, then Y  X also holds in R. • Disprove that if X, Y and Z are sets of attributes of a relational schema R, and the fds X  Y and Y  Z hold in R, then Z  X also holds in R. • Disprove that if X, Y and Z are sets of attributes of a relational schema R, and the fd XY  Z holds in R, then X  YZ also holds in R.

20. Bogus rules 3 • Disprove that if X and Y are sets of attributes of a relational schema R, and the fd X  Y holds in R, then Y  X also holds in R. • Solution: • Consider the following relation instance, • where we use singletons for X and Y: • We see that X  Y holds, but not Y  X

21. Bogus rules 4 • Disprove that if X, Y and Z are sets of attributes of a relational schema R, and the fds X  Y and Y  Z hold in R, then Z  X also holds in R. • Solution: • Consider the following relation instance, • where we use singletons for X, Y, and Z: • We see that both X  Y and Y  Z hold • But not Z  X.

22. Bogus rules 5 • Disprove that if X, Y and Z are sets of attributes of a relational schema R, and the fd XY  Z holds in R, then X  YZ also holds in R. • Solution: • Consider the following relation instance • where we use singletons for X, Y, and Z: • We see that XY  Z holds, but not X  YZ.

23. Summary • We have learned how to prove & disprove FDs based on the definition

24. … to follow Functional Dependencies (FDs) applied (2)

25. Questions?