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Here's the code that detects left recursion:

The function walks the grammar tree and if it loops past a given
threshold, an error is raised.

LPeg could support left-recursion. Sérgio Medeiros, Fabio Mascarenhas
and Roberto Ierusalimschy wrote a paper that describes an extension to
the engine, but they did not publish the code, for some reason.

Rostislav Sacek added left recursion support in LPegLJ, but it only
works with LuaJIT.

—Pierre-Yves Gérardy

On Mon, Dec 15, 2014 at 4:53 PM, Sam Putman <> wrote:
> On Sun, Dec 14, 2014 at 10:56 PM, Sean Conner <> wrote:
>>   LPeg is a Lua implementation of Parsing Expresson Grammars, which are
>> *NOT* (I repeat, *NOT*) the same as LR() or LL() parsers.  There's a bit
>> more information about this on the Wikipedia:
>>   -spc
> Absolutely correct, now, let's focus.
> The lpeg engine has to detect potential left recursion, and reject it.
> How does it do this?
> It looks at a call to "patt", and sees whether another call to "patt" can be
> reached without consuming any input.  We check the rules, one at a time, to
> assure that the second "patt" call either consumes or fails before it
> reaches a third.
> The only type of capture which can modify this *logic* in a way that *still
> prevents infinite loops* is #patt, and only under this deterministic
> circumstance: If we have a pattern 'token', a looping pattern "expr", and a
> calling pattern "factor", then if factor looks like "factor = #token * expr"
> and expr looks like "expr = token + factor", we're okay.
> Specifically, any #patt encountered before a potential left-recursion should
> be followed to see if a later call to `patt` is encountered. This must
> happen before the cursor moves, and before the recursive call to the
> "factor" type rule.
> Think of #patt as a promise to the grammar compiler that 'patt' will be
> encountered. The compiler may statically confirm that this happens.
> Currently, it treats #patt as any other pattern that doesn't consume input,
> but in fact, it adds a new constraint which may be propagated.
> One more time: I am making a logical claim about an algorithm. I would like
> to see demonstration code, and I'll tell you if the algorithm would say
> "cannot loop infinitely" or "might loop infinitely". I am especially
> interested in grammars for which these categories cannot be assigned: I
> haven't proven that it's impossible, because I don't have the time, but I
> suspect it is.