Catch strings with equal a's and b's (only a and b).

Equal-count language; aab has 2a/1b; hard without counting.

@regexwizard⚡⚡⚡⚡⚡PAR 32Mixed11 players · 10 solves · 7 views

About this hunt

5 words to catch, 5 words to spare, plus hidden checks. It mixes several techniques. Added 14 Jun by @regexwizard.

null_ptr· 21 Jun
par 32?? ok the d5 is not a bluff. is this even possible without recursion
- regexwizardAuthor· 21 Jun
  it is, just barely. a lookahead can count a's and b's in lockstep. it's not pretty.
  - null_ptr· 22 Jun
    'not pretty' is doing heavy lifting there lol
vex· 23 Jun
wait — abab AND baab are both targets but aab is innocent? ohhh it's equal COUNTS, order doesn't matter. completely changes how I was thinking about this.
greedy_star· 24 Jun
spent a solid half hour certain this was impossible ("regex can't count!") before remembering lookaheads are a thing. humbled.
pat_match· 26 Jun
the aaab innocent is the gut-check — three a's, one b, not equal. easy to slip through if you only test "has both letters".
slate_otter· 27 Jun
anyone actually near par on this? best I've got is 40-something chars and it looks like keyboard smashing
- byte_smith· 27 Jun
  mine's 38 and equally cursed. i don't think clean is an option here honestly
GageMod· 24d
I'm a wild horse running on the road