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As GrandFather said, your regex is equivalent to /a*b/, so it's easy to see that the regex can never match. The slowdown is because of the backtracking done by Perl's regex engine.

Let's look at a smaller example:

"aaaa" =~ /a*a*b/
Except I'll make the following changes:
  • Instead of /b/ at the end, I'll use /(?!)/, which does the same thing here (it is a regex that always fails, just like the /b/ will always fail to match anywhere in this particular string).1
  • I'll capture both /a*/'s so I can look at what the regex engine puts in them
  • Before it gets to the end where it always fails (the /(?!)/ part), I'll have it print what parts of the string it tried matching to the /a*/'s.
The result is this:
"aaaa" =~ m/ (a*) (a*) (?{ print "Tried $`($1)($2)\n"; }) (?!) /x;
The output:
Tried (aaaa)() Tried (aaa)(a) Tried (aaa)() ... ... Tried aaa()(a) Tried aaa()() Tried aaaa()()
So you can see that it tries a way to fill $1 and $2 (and $`), fails to match /(?!)/, backtracks, and tries again repeatedly. It tries every possible way before the regex match operation finally returns with a failure. In this small example it tries 35 combinations before eventually failing.

Now if you make the string a lot longer, and give it more ways for the prefixes to match (more /a*/ regexes to fill), it will take a whole lot longer. By my calculations, the regex you gave above will try

24670925422945900903156716 (= 2e25)
combinations before eventually failing. Even at a billion combinations per second, that's still 782 million years ;)

1: If I left /b/ as the last part of the regex instead of /(?!)/, then the regex engine seemed to optimized away my print statement.

Update: revised my big number calculation, not that it makes a huge difference. It's 61+33 choose 33 if you're curious. That's the number of ways to put 61 balls (the a's) into 34 ordered bins (32 /a*/ regexes, plus $` and $').

blokhead


In reply to Re: slow regular expressions by blokhead
in thread slow regular expressions by jesuashok

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