[d66]

interesting!

what would [ab]* be? for computing an ordinal number the only real difficulty is how to handle kleene star: given ord(X) how do we calculate ord(X*)?

but as you probably noticed i'm a bit out of my depth when dealing with ordinals.

[contravariant]

Unless I'm very mistaken adding onto the end of a string doesn't affect lexicographic order, so that's effectively [ab]*. The ordinality of [ab] is simple, it's 2, but the kleene star is a bit of an odd one, it's not quite exponentiation.

To reason about the kleene star it's a bit simpler to consider something like R^*n, where you repeat up to n times. Obviously R^*0 = 1 and R^*S(n) can be built from R^*n by picking an element of R^*n and appending either nothing or an element of R, here the element of R^*n determines most of the order and 'nothing' orders in front. For technical reasons the corresponding ordinal is (1+R) R^*n, which is backwards from how you'd expect it and how you'd normally define exponentiation.

The kleene star can be recovered by taking the limit, identifying R^*n with it's image in R^*S(n). Which also doesn't quite work as nicely as you'd hope (normally the image is just a downward closed subset, it's not in this case).

I think [ab]* is equivalent to something like the rational part of the Cantor set. Not sure if there's a simpler way to describe it, it's nowhere near as simple as 2^ω, which is just ω.

[contravariant]

Hmm, turns out this fails to be an ordinal because regular languages aren't well-ordered (except if you reverse the lexicographic order, maybe?). They are what is called an order type, and it looks like it should be possible to identify them with subsets of the rationals, if you wanted to.

Perhaps reversing the lexicographic order makes more sense, in that case longer tuples simply order last so R^* = 1 + R + R^2 + ..., the limit here is much easier since R^*n = 1 + R + ... + R^n is downwards closed as a subset of R^*S(n).

Then again in that scenario [ab]* is simply ω because it is effectively the same as just writing an integer in binary, so it is less interesting in a way.

[skulk]

I would say [ab]* has order type omega. That's because the set of strings it represents is: {epsilon, ab, abab, ababab, abababab, ...} which looks a lot like omega. (epsilon = empty string)

What this exercise really is is finding a canonical way to order a regular language (the set of strings a regexp matches). For example, a*b* could be {epsilon, a, aa, aaa, aaa, aaaa, ..., b, bb, bbb, bbbb, bbbbb, bbbbbb, ..., ab, aab, aaab, aaaab, ..., abb, aabb, ..., ...} which looks a lot like omega ^ 2 (not 2 * omega like I said before). However, you could also re-arrange the set to look like omega: {epsilon, a, b, aa, bb, ab, aaa, bbb, aab, abb, bbb, ...} (strings of length 1, length 2, length 3, etc)

I propose the following: for any two strings in the regular language, the one that comes first is the one whose kleene-star repetition counts come first lexicographically. More concretely, for the language a*b*, aaaab represents a repetition count of (4, 1) which and bbb represents (0, 3). (0, 3) comes before (4, 1) lexicographically, so bbb comes before aaaab. This admits the following ordering: {epsilon, b, bb, bbb, bbbb, ..., a, abb, abbb, abbbb, ..., aa, aab, aabb, aabbb, ..., ...} which is omega ^ 2 which "feels" right to me. Another rule is for the regular language (X|Y) and two strings x from X and y from Y, x should always come before y in our ordered set representation.

Hold on, what about nested kleene-stars? (a*)* is just a*, but (a*b*)* is distinct from a*b*. However, the "kleene-star counts" analysis from above breaks down because there are now multiple ways to parse strings like aab. I don't really know how to classify these regular languages as ordinals yet.

I don't really see any useful applications of this, but it's still fun to think about. The game I'm playing is thinking about a random ordinal and trying to come up with a regular language that, under my ordering rule above, looks like that ordinal. Let's try 2 * omega (which looks like this: {0, 1, 2, 3, 4, 5, ..., omega, omega + 1, omega + 2, omega + 3, ...} e.g. 2 copies of omega "concatenated"):

a*|b* = {epsilon, a, aa, aaa, aaaa, aaaaa, ..., b, bb, bbb, bbbb, ...} => 2 * omega.

Some more examples:

omega ^ 3: a*b*c*

omega ^ 2 + omega: a*b*|c*

Maybe we can write down some composition rules:

let X and Y be regular languages and ord(X) and ord(Y) be their ordinal representations. Then,

X|Y => ord(X) + ord(Y)

XY => ord(X) * ord(Y)

X* => ord(X) * omega

I haven't checked if these actually work, this is just a long rambly comment of dubious mathematical value.

[wnoise]

[ab] is a character class, not a parenthesized catenation. It's (a|b), not (ab). So [ab]* is {epsilon, a, b, aa, ab, ba, bb, ...}