> search a document for a pattern and it takes a second. search one a hundred times larger and it doesn't take a hundred seconds - it can take almost three hours.

Most of this is about quadratic time find-all operations where a search operation is linear. But it's also still possible to get quadratic behaviour out of a single search without catastrophic backtracking, more easily than you might expect. In late January to early February, Tim Peters was talking about an example of this on the Python forums (see e.g. https://discuss.python.org/t/add-re-prefixmatch-deprecate-re...) and also related the experience of trying to diagnose the issue with AI (see https://discuss.python.org/t/claude-code-how-much-hype-how-m... and onward). Peters' example was:

  \d+\s+

on a string containing only digits, a prefix match takes O(n) time as it considers every possible end position for the digit, and immediately sees no following whitespace. But the search is quadratic because it has to repeat that O(n) work at every position; the regex engine can't track the fact that it's already examined the string and found no whitespace, so it re-tries each digit match length.

(This is arguably "backtracking" since it tries the longest match first, but clearly not in a catastrophic way; if you use `\d+?` instead then of course it only searches forward but is still O(n). It actually is slower in my testing in the Python implementation; I don't exactly know why. As noted in the discussion, the possessive quantifier `\d++` is considerably faster, and of course doesn't backtrack, but still causes O(n^2) searching. The repeated attempts to match `\s+` aren't the problem; the problem is repeatedly looking for digits in places where digits were already found and rejected.)

The way to fix this proposed in the discussion is to use a negative lookbehind assertion before the digits: `(?<!\d)\d+\s+`. This way, the regex engine can bail out early when it's in the middle of a digit string; if the previous character was a digit, then either `\d+\s+` doesn't match here, or it would have matched there.

A simpler idea is to just search for `\d\s+`, or even `\d\s` — since these will be present if and only if `\d+\s+` is. This way, though, you still need to do extra work with the partial match to identify the start and end of the full match. My first idea was to use positive* lookbehind for the digits, since the lookbehind match doesn't need to backtrack. In fact lookbehinds require a fixed-length pattern, so this is really just a more complicated way to do the `\d\s+` simplification.

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> Hyperscan (and its fork Vectorscan) is a true linear-time all-matches regex engine. it achieves this by using "earliest match" semantics - reporting a match the moment the DFA enters a match state, instead of continuing to find the longest one.

Is this not just equivalent to forcing "reluctant" quantifiers (`\d+?`) everywhere?

Finding all regex matches has always been O(n²)