It goes from start of the first match to the longest "alive" end, in practice it will go to a dead state and return after finding the match end.

there's an implicit `.*` in front of the first pass but i felt it would've been a long tangent so i didn't want to get into it.

so given input 'aabbcc' and pattern `b+`,

first reverse pass (using `.*b+`) marks 'aa|b|bcc'<-

the forward pass starts from the first match:

'aa->b|b|cc' marking 2 ends

then enters a dead state after the first 'c' and returns the longest end: aa|bb|cc

i hope this explains it better

▲

masfuerte 2 hours ago | parent [-]

Cheers. I was more confused by how you were doing multiple matches. So I read the paper, which describes the AllEnds algorithm. If I understand correctly, the reverse pass captures all of the match starts and these need to be remembered for the forward pass. Which is what you were showing above, but I didn't follow it.

So, once it gets going, a traditional engine can produce matches iteratively with no further allocation, but RE# requires allocation proportional to the total number of matches. And in return, it's very much faster and much easier to use (with intersection and complement).

	▲	ieviev 2 hours ago \| parent [-]
		Yes, exactly correct It's also beneficial to merge some of the matching locations into ranges where possible, so when `a*` matches a long sequence of '\|a\|a\|a\|a\|a\|', it can be represented as a range of (0,5), we do this to keep the lookaround internal states smaller in the engine.