Re^3: check for square-number with a regex

These regexes are much more powerful than regular expressions, in fact they can do things that even Typ-1 langauges (context-free languages) can't. I've not yet seen a proof that they are Turing complete, so I don't know if they are.

In a sense they're ‘obviously’ Turing-complete, since they can contain embedded Perl code; but I assume that's not what you meant. What's another example of Perl regexes recognising a language that a CFG can't? (It sounds vaguely like a challenge, but I'm just politely curious.)

Comment on Re^3: check for square-number with a regex

Replies are listed 'Best First'.
Re^4: check for square-number with a regex by moritz (Cardinal) on Oct 23, 2009 at 15:47 UTC
First of all sorry for getting the numbering wrong, I misremembered REG as being Typ-0 and Turing as being Typ-3, it's the other way round. What's another example of Perl regexes recognising a language that a CFG can't? `m/^(\w+)\1$/` Is already outside CFG. CFG is parsed by a non-deterministic finite automaton which has access to a stack of infinite size. Since it can only access the top element of the stack, it has only access to the capture in reverse order, or with a finite look-behind on the stack (implemented by storing some values in the state of the NFA). A more formal proof can be made with the Pumping lemma for context-free languages. I also think that the regex primality test is outside of CFG, but I've never proven that, so don't rely on it ;-)	[reply] [d/l]
Re^5: check for square-number with a regex by JadeNB (Chaplain) on Nov 26, 2009 at 23:35 UTC
I also think that the regex primality test is outside of CFG, but I've never proven that, so don't rely on it ;-) Doesn't the pumping lemma that you cited also show what you claim? Specifically, if there were a CFG that recognised exactly the strings of `1`s of prime length, then we would have the assertion that any sufficiently large prime `s` could be written as `s = u + v + x + y + z`, with everyone non-negative and `v + y` at least `1`, in such a way that `u + x + i(v + y) + z` was also prime for every non-negative `i`. However, in that case, if `u + x + z = 0`, then we can take `i` to be composite, so that `u + x + z + i(v + y) = i(v + y)` is composite; or otherwise, we may take `i = (v + y + 2)(u + x + z)`, so that `u + x + z + i(v + y) = (v + y + 1)^2(u + x + z)` is composite. This is a contradiction. UPDATE: Forgot to handle the case `u + x + z = 0`. UPDATE 2: Oops, and I was quite sloppy in my choice of `i` even when `u + x + z` is positive. I think that it's OK now.	[reply] [d/l] [select]