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Lecture 6: Schema refinement: Functional dependencies

Lecture 6: Schema refinement: Functional dependencies. www.cl.cam.ac.uk/Teaching/current/Databases/. Next two lectures. Recall: Database design lifecycle. Requirements analysis User needs; what must database do? Conceptual design High-level description; often using E/R model

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Lecture 6: Schema refinement: Functional dependencies

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  1. Lecture 6: Schema refinement: Functional dependencies www.cl.cam.ac.uk/Teaching/current/Databases/

  2. Next two lectures Recall: Database design lifecycle • Requirements analysis • User needs; what must database do? • Conceptual design • High-level description; often using E/R model • Logical design • Translate E/R model into relational schema • Schema refinement • Check schema for redundancies and anomalies • Physical design/tuning • Consider typical workloads, and further optimise

  3. Today’s lecture • Why are some designs bad? • What’s a functional dependency? • What’s the theory of functional dependencies? • (Next lecture: How can we use this theory to classify redundancy in relation design?)

  4. Not all designs are equally good • Why is this design bad? Data(sid,sname,address,cid,cname,grade) • Why is this one preferable? Student(sid,sname,address) Course(cid,cname) Enrolled(sid,cid,grade)

  5. An instance of our bad design

  6. Evils of redundancy • Redundancy is the root of many problems associated with relational schemas • Redundant storage • Update anomalies • Insertion anomalies • Deletion anomalies • LOW TRANSACTION THROUGHPUT • In general, with higher redundancy, if transactions are correct (no anomalies), then they have to lock more objects thus causing greater contention and lower throughput • (Aside: Could having a dummy value, NULL, help?)

  7. Decomposition • We remove anomalies by replacing the schema Data(sid,sname,address,cid,cname,grade) withStudent(sid,sname,address) Course(cid,cname) Enrolled(sid,cid,grade) • Note the implicit extra cost here • Two immediate questions: • Do we need to decompose a relation? • What problems might result from a decomposition?

  8. Functional dependencies • Recall: • A key is a set of fields where if a pair of tuples agree on a key, they agree everywhere • In our bad design, if two tuples agree on sid, then they also agree on address, even though the rest of the tuples may not agree

  9. Functional dependencies cont. • We can say that sid determines address • We’ll write thissid  address • This is called a functional dependency (FD) • (Note: An FD is just another integrity constraint)

  10. Functional dependencies cont. • We’d expect the following functional dependencies to hold in our Student database • sid  sname,address • cid  cname • sid,cid  grade • A functional dependency X  Y is simply a pair of sets (of field names) • Note: the sloppy notation A,B  C,D rather than {A,B}  {C,D}

  11. Formalities • Given a relation R=R(A1:1, …, An:n), and X, Y ({A1, …, An}), an instance r of R satisfies XY, if • For any two tuples t1, t2 in R, if t1.X=t2.X then t1.Y=t2.Y • Note: This is a semantic assertion. We can not look at an instance to determine which FDs hold (although we can tell if the instance does not satisfy an FD!)

  12. Properties of FDs • Assume that X  Y and Y  Z are known to hold in R. It’s clear that X  Z holds too. • We shall say that an FD set F logically implies X  Y, and write F [X  Y • e.g. {X  Y, Y  Z} [X  Z • The closure of F is the set of all FDs logically implied by F, i.e. F+@ {XY | F [XY} • The set F+ can be big, even if F is small 

  13. Closure of a set of FDs • Which of the following are in the closure of our Student FDs? • addressaddress • cidcname • cidcname,sname • cid,sidcname,sname

  14. Candidate keys and FDs • If R=R(A1:1, …, An:n) with FDs F and X{A1, …, An}, then X is a candidate key for R if • X  A1, …,An  F+ • For no proper subset YX is Y  A1, …,An  F+

  15. Armstrong’s axioms • Reflexivity: If YX then F \ XY • (This is called a trivial dependency) • Example: sname,addressaddress • Augmentation: If F \ XY then F \ X,WY,W • Example: As cidcname then cid,sidcname,sid • Transitivity: If F \ XY and F \ YZ then F \ XZ • Example: As sid,cidcid and cidcname, then sid,cidcname

  16. Consequences of Armstrong’s axioms • Union: If F \ XY and F \ XZ then F \ XY,Z • Pseudo-transitivity: If F \ XY and F \ W,YZ then F \ X,WZ • Decomposition: If F \ XY and ZY then F \ XZ Exercise: Prove that these are consequences of Armstrong’s axioms

  17. Proof of Union Rule Suppose that F \ XY and F \ XZ. By augmentation we have F \ XX,Y since X U X = X. Also by augmentation F \ X,YZ,Y Therefore, by transitivity we have F \ XZ,Y QED

  18. Functional Dependencies Can be useful in Algebraic Reasoning Suppose R(A,B,C) is a relation schema with dependency AB, then (This is called Heath’s rule.)

  19. Proof of Heath’s Rule First show that Suppose then and Since we have

  20. Proof of Heath’s Rule (cont.) In the other direction, we must show that Suppose Then there must exist records and There must also exist so that But the functional dependency tells us that QED Therefore, we have

  21. Equivalence • Two sets of FDs, F and G, are said to be equivalent if F+=G+ • For example: {(A,BC), (AB)} and {(AC), (AB)} are equivalent • F+ can be huge – we’d prefer to look for small equivalent FD sets

  22. Minimal cover • An FD set, F, is said to be minimal if • Every FD in F is of the form XA, where A is a single attribute • For no XA in F is F-{XA} equivalent to F • For no XA in F and ZX is (F-{XA}){ZA} equivalent to F • For example, {(AC), (AB)} is a minimal cover for {(A,BC), (AB)}

  23. More on closures • FACT: If F is an FD set, and XYF+ then there exists an attribute AY such that XAF+

  24. Why Armstrong’s axioms? • Soundness • If F \ XY is deduced using the rules, then XY is true in any relation in which the dependencies of F are true • Completeness • If XY is is true in any relation in which the dependencies of F are true, then F \ XY can be deduced using the rules

  25. Soundness • Consider the Augmentation rule: • We have XY, i.e. if t1.X=t2.X then t1.Y=t2.Y • If in addition t1.W=t2.W then it is clear that t1.(Y,W)=t2.(Y,W)

  26. Soundness cont. Consider the Transitivity rule: • We have XY, i.e. if t1.X=t2.X then t1.Y=t2.Y (*) • We have YZ, i.e. if t1.Y=t2.Y then t1.Z=t2.Z (**) • Take two tuples s1 and s2 such that s1.X=s2.X then from (*) s1.Y=s2.Y and then from (**) s1.Z=s2.Z

  27. Completeness • Exercise • (You may need the fact from slide 23)

  28. Attribute closure • If we want to check whether XY is in a closure of the set F, could compute F+ and check – but expensive  • Cheaper: We can instead compute the attribute closure, X+,using the following algorithm: • Then F \ XY iff Y is a subset of X+ Try this withsid,snamecname,grade closure:= X; repeat until no change{ ifUVF, where Uclosure then closure:=closureV };

  29. Preview of next lecture: Goals of normalisation • Decide whether a relation is in “good form” • If it is not, then we will “decompose” it into a set of relations such that • Each relation is in “good form” • The decomposition has not lost any information that was present in the original relation • The theory of this process and the notion of “good form” is based on FDs

  30. Summary You should now understand: • Redundancy and various forms of anomalies • Functional dependencies • Armstrong’s axioms Next lecture: Schema refinement: Normalisation

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