1. Functional Dependencies

2. Babies Exercise 2.2.5: At a birth, there is one baby (twins would be represented by two births), one mother, any number of nurses, and any number of doctors. Suppose, therefore, that we have entity sets Babies, Mothers, Nurses, and Doctors. Suppose we also use a relationship Births, which connects these four entity sets. Note that a tuple of the relationship set for Births has the form (baby, mother, nurse, doctor)

3. Babies (Cont’ed) There are certain assumptions that we might wish to incorporate into our design. For each, tell how to add arrows or other elements to the E/R diagram in order to express the assumption. a) For every baby, there is a unique mother. b) For every combination of a baby, nurse, and doctor, there is a unique mother. c) For every combination of a baby and a mother there is a unique doctor.

4. Functional Dependencies X -> A is an assertion about a relation R that whenever two tuples of R agree on all the attributes of X, then they must also agree on the attribute A. –Say “X -> A holds in R.” –Convention: …, X, Y, Z represent sets of attributes; A, B, C,… represent single attributes. –Convention: no set formers in sets of attributes, just ABC, rather than {A,B,C}.

5. Let’s consider the relation: –Movie(title, year, length, filmType, studioName, starName). There are several functional dependencies that we can reasonably assert. title year  length title year  filmType shorthand: title year  length filmType studioName title year  studioName These assertions make sense if we remember the original design: –Attributes title and year form a key for movie objects. –Thus, given a title and a year there is a unique length, filmType and a unique owning studio. Functional Dependencies - Example

6. Example Drinkers(name, addr, beersLiked, manf, favBeer) Reasonable FD’s to assert: 1.name -> addr 2.name -> favBeer 3.beersLiked -> manf

7. Example Data nameaddr beersLiked manffavBeer JanewayVoyager Bud A.B.WickedAle JanewayVoyager WickedAle Pete’sWickedAle SpockEnterprise Bud A.B.Bud Because name -> addr Because name -> favBeer Because beersLiked -> manf

8. FD’s With Multiple Attributes No need for FD’s with > 1 attribute on right. –But sometimes convenient to combine FD’s as a shorthand. –Example: name -> addr and name -> favBeer become name -> addr favBeer > 1 attribute on left may be essential. –Example: bar beer -> price

9. K is a superkey for relation R if K functionally determines all of R. K is a key for R if K is a superkey, but no proper subset of K is a superkey. Example. Attributes {title, year, starName} form a key for the previous Movie relation. Why? Keys of Relations

10. Because the first property of keyness is satisfied: –title year functionally determines length, filmType and studioName. So does their superset {title, year, starName}. –starName functionally determines itself (starName). So does the superset {title, year, starName}. –Concluding title, year, starName functionally determine all the attributes. And, also the second property of keyness is satisfied: No proper subset of {title, year, starName} functionally determines all attributes. –{title, year} do not determine starName, because a movie can have many stars. –{year, starName} do not determine title, because we can have a star in two movies in the same year. –{title, starName} do not determine let’s say year, because we could have two movies with the same title made in different years. It can be the case these two movies to have a star in common. It is very difficult to find such cases. However, if we make {title, starName} key, then we prohibit the user to insert a new movie, which has the same title as a previous inserted old movie, and where a star plays again (not necessary in the same role). Remember, the constraints are supposed to hold for all the instances, not only for the current DB instance. Keys of Relations (Continued)