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

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2 … previous Generalities on Databases

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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

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4 Content 1.Generalities DB 2.Integrity constraints (FD revisited) 3.Relational Algebra (revisited) 4.Query optimisation 5.Tuple calculus 6.Domain calculus 7.Query equivalence 8.LLJ, DP and applications 9.Temporal Data 10.The Askew Wall

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5 Functional Dependency -A functional dependency (FD) has the form X Y where 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 t 1 and t 2 of r t 1 (X) = t 2 (X) t 1 (Y) = t 2 (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])

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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). XY means X Y. (***) Should actually be X Attr(R) (heading)

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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

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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.

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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!

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10 Ex: Augmentation and Transitivity rules 1.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. 2.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.

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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) (C D) (A C) (B D)) ( r R, t1, t2 r : t1[X]=t2[X] t1[Z]=t2[Z] : t1[Y]=t2[Y] 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)

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

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13 Augumentation (formal -1) (1)( r R, t1, t2 r ::t1[X]=t2[X] t1[Y]=t2[Y]) (this is the definition of X Y) (2) ( r R, t1, t2 r :: t1[Z]=t2[Z] t1[Z]=t2[Z]) (this is always true) (3)Since both (1) and (2) hold, we can conjugate them: ( r R, t1, t2 r :: t1[X]=t2[X] t1[Y]=t2[Y]) ( r R, t1, t2 r :: t1[Z]=t2[Z] t1[Z]=t2[Z]) (domain splitting) (4)( r R, t1, t2 r :: (t1[X]=t2[X] t1[Y]=t2[Y]) t1[Z]=t2[Z] t1[Z]=t2[Z]))

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14 Augumentation (formal -2) (domain splitting) (4)( r R, t1, t2 r :: (t1[X]=t2[X] t1[Y]=t2[Y]) t1[Z]=t2[Z] t1[Z]=t2[Z])) (because of Lemma 1: ((A B) (C D)) ((A C) (B D))) ( r R, t1, t2 r :: (t1[X]=t2[X] t1[Z]=t2[Z]) (t1[Y]=t2[Y] 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)

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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])) ( 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]) (t1[Y]=t2[Y] t1[Z]=t2[Z]))

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16 Transitivity (2) (domain splitting) ( r R, t1, t2 r ::( t1[X]=t2[X] t1[Y]=t2[Y]) (t1[Y]=t2[Y] t1[Z]=t2[Z])) (because of Lemma 2: ((A B) (B C)) (A C)) ( r R, t1, t2 r :: t1[X]=t2[X] t1[Z]=t2[Z]) (this is the definition of X Z)

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17 Lemma 2 ((A B) (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 ~Y), distribute negation over conjunction) (A ~B) v (B ~C) v (~A v C) (use ((X ~Y) v ~X) (~Y v ~X)) ~A v ~B v B v C ( (X v ~X) true; true/false elimination) true

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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.

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19 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. 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. 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.

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20 Bogus rules 3 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 XY 10 00

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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. XYZ 000 100

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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. XYZ 000 010

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

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

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25 Questions?

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