3.4 Modeling with relations  (Page 2/2)

We can use the binary relation $\mathrm{thinksIsYummy}$ : In particular, $\mathrm{thinksIsYummy}(\text{Ian}, \text{anchovies})=\mbox{false}$ but $\mathrm{thinksIsYummy}(\text{Phokion}, \text{anchovies})=\mbox{true}$ What set are we using, as the domain for this? Really, the domain is the union of people and pizza-toppings.So $\mathrm{thinksIsYummy}(\text{radishes}, \text{brusselsSprouts})$ is a valid thing to write down; it would be false. Note that if working with such a domain,having unary predicates $\mathrm{isVegetable}$ and $\mathrm{isPerson}$ would be useful.

The proposed formula asserts that each pair has been in some movie together,but they each could have been different movies without being in the same one simultaneously.As a counterexample, it is true that $\mathrm{hasStarredWith}(\text{Charlie Chaplin}, \text{Norman Lloyd})$ (as witnessed by Limelight , 1952), $\mathrm{hasStarredWith}(\text{Norman Lloyd}, \text{Janeane Garofolo})$ (as witnessed by The Adventures of Rocky and Bullwinkle , 2000), and if we generously include archive footage, $\mathrm{hasStarredWith}(\text{Charlie Chaplin}, \text{Janeane Garofolo})$ (as witnessed by Outlaw Comic: The Censoring of Bill Hicks , 2003); however, they have not all been in a movie together.Might the counterexample you chose become nullified, in the future?

As always, there are several ways of modeling this problem. We'll outline three.

First, we could augment the $\mathrm{hasStarredWith}$ to be a ternary (3-input) relation to include the movie. Like in the $\mathrm{yummy}$ extension , the domain would then include both actors and movies, and we'dalso want relations to know which is which.

Second, we could use a bunch of relations. Starting with the familiar binary $\mathrm{hasStarredWith}$ , we'd add the ternary $\mathrm{hasStarredWith3}$ , the quaternary $\mathrm{hasStarredWith4}$ , …. Our domain would just be actors.However, we'd either need an infinite number of such relations, which we normally don't allow, or we'd need an arbitrary capon the number of people we're interested in at a time.

Third, we could use sets of actors, instead of individuals.We'd need only one relation, $\mathrm{haveStarredtogether}$ , that states a set of actors have starred together in a single movie.

This is effectively Disjunctive or Conjunctive Normal Form, limited to clauses of one term each.

Yes. Two examples are $\neg ((\mathrm{genre}=\mathrm{Classical})\lor (\mathrm{genre}=\mathrm{Holiday}))$ , and $(\mathrm{genre}=\mathrm{Rock})\land (\mathrm{Rating}\ge 4)\lor (\mathrm{genre}=\mathrm{Classical})$ .

For the first example, $\neg ((\mathrm{genre}=\mathrm{Classical})\lor (\mathrm{genre}=\mathrm{Holiday}))$ , we can clearly use DeMorgan's law and make the query $\neg (\mathrm{genre}=\mathrm{Classical})\land \neg (\mathrm{genre}=\mathrm{Holiday})$ .

However, for $(\mathrm{genre}=\mathrm{Rock})\land (\mathrm{Rating}\ge 4)\lor (\mathrm{genre}=\mathrm{Classical})$ there is no equivalent one-term-per-clause DNF or CNF formula! Budding logicians might wonder how you actually prove this claim!

Fortunately, iTunes has a way around this. Playlist membership or non-membership is itself an available predicate, allowing you to nest playlists. Thus, you can build a playlist $\mathrm{GoodOrClassical}$ for $(\mathrm{Rating}\ge 4)\lor (\mathrm{genre}=\mathrm{Classical})$ , then another $(\mathrm{genre}=\mathrm{Rock})\land \mathrm{GoodOrClassical}$ for the desired result.