Nullability (Regular Expressions)

Question

In Brzozowski's "Derivatives of Regular Expressions" and elsewhere, the function δ(R) returning λ if a R is nullable, and ∅ otherwise, includes clauses such as the following:

δ(R1 + R2) = δ(R1) + δ(R2)
δ(R1 · R2) = δ(R1) ∧ δ(R2)

Clearly, if both R1 and R2 are nullable then (R1 · R2) is nullable, and if either R1 or R2 is nullable then (R1 + R2) is nullable. It is unclear to me what the above clauses are supposed to mean, however. My first thought, mapping (+), (·), or the Boolean operations to regular sets is nonsensical, since in the base case,

δ(a) = ∅ (for all a ∈ Σ)
δ(λ) = λ
δ(∅) = ∅

and λ is not a set (nor is a set the return type of δ, which is a regular expression). Furthermore, this mapping isn't indicated, and there is a separate notation for it. I understand nullability, but I'm lost on the definition of the sum, product, and Boolean operations in the definition of δ: how are λ or ∅ returned from δ(R1) ∧ δ(R2), for instance, in the definition off δ(R1 · R2)?

Answer 1

I think you were right to map + and ^ to boolean or and and respectively. It looks like the two lines you cited deal with alternation (sum) and concatenation (product):

δ(R1 + R2) = δ(R1) + δ(R2)

The alternation of R1 and R2 is nullable if R1 is nullable, R2is nullable, or both R1 and R2 are nullable.

δ(R1 · R2) = δ(R1) ∧ δ(R2)

The concatenation of R1 and R2 is only nullable if both R1 and R2 are nullable.

See here for an Haskell implementation of these rules.

Answer 2

(I can't look into the article by Brzozowski in order to understand better what is meant there), but I can suggest 2 ways to interprete this notation (apart from gettingalong with the notation, I see, there is no question: the intended sense of this definition is well understood):

1) On left of the definition, we have just "syntactic" patterns for the regular expressions. On the right, we produce sets; remember, that a regular expression is a way to denote a language (a set), and so this way to write down the definition becomes understandable: on the right, we simply use some (simple) regular expressions as a short way to refer to sets. I.e.,∅ means the empty language (the empty set), and λ (if interpreted as a regular expression) means the language containing just the empty word (the set with this element).

The operations are simply operations on sets: probably union, and intersection.

If the notation is interpreted this way, there is no contradiction with the used notation to defian the base case: again, "a" is a regular expression which stands there to mean the language with the word "a".

2) We build regular expressons on the right, in the first place, but the author has extended the operations which build regular expressions with the wedge, which has the semantics of intersection of languages.

Answer 3

I think you're getting caught out by the notational liberties taken by the author. The return type of δ(R) is most certainly a set, or rather a language. If you look at the definition:

you can see that there is an inconsistency in the return type, formally λ is an element, but ∅ is the empty language... What it should say is:

The fact that the author uses λ for both the empty string as well as the language containing only the empty string is further evidenced by his definition of the Kleene star operator:

Clearly, the last part should be if we want to be pedantic.

Given that the return type of δ(R) is a set, or rather a language, the equations you give make perfect sense and express exactly what you described.