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Syntactic Confectionery Delight

Re: Pig Latin

by davisagli (Scribe)
on Jun 19, 2001 at 18:41 UTC ( #89649=note: print w/replies, xml ) Need Help??

in reply to Pig Latin

I had some fun with this. My shortest fully-functional attempt is this 60-character regexp:


with these features:

  • Handles multiple consonants at start of word (and handles 'qu' correctly)
  • Correctly handles y-related idiosyncrasies: yummy becomes ummyyay, but yttrium becomes yttriumway and rhythm becomes ythmrhay
  • Handles numbers correctly (42 doesn't become 42ay)
  • Counts apostrophe as part of word, so "don't" becomes "on'tday" (and handles other punctuation correctly)

Note that if we ignore two words (yttrium and ytterbium), we can safely bring it down to 52 chars:


And if we decide to switch to the dialect of pig latin that doesn't add 'w' on vowel-words, it's down to 44:


If I combine that with perlmonkey's attempt (losing a bit of functionality in the process, although it still works pretty well) I can reach 36:


Also note that my (un-golfed) attempt at a pig latin converter in Visual Basic took up almost 3500 characters (!) without handling nearly all of the exceptions mentioned here. This is (one of the many reasons) why I love Perl!

Thanks to everyone for their previous attempts, which helped me quite a bit. Comments are welcome.


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[hippo]: Understood. I'll have to go through the code and see if it's doing anything fancy with ties, dual-vars or non-scalars. In the end, it's probably a bug though.
[Corion]: Aaah - you should be able to do this with overload, but I would hit somebody really hard if they constructed objects that are true but the empty string, and you not knowing about the domain knowledge where this makes sense
[Eily]: you could tie a variable into not having the same value each time, if you like to make people who try to debug your code facepalm
[Corion]: perl -wle 'package o; use overload q("") => sub {warn "str"; ""}, bool => sub{warn "bool"; 1}; package main; my $o={}; bless $o => o; print "Yay" if ($o && !length($o))'
[Corion]: But people writing such code should document the objects they construct and why it makes sense for an object to be invisible as string while being true in a boolean context
[hippo]: That's equal parts clever and horrendous.
[Eily]: the overload version wouldn't return true with "$x" && !length $x though, I guess
[hippo]: The more I look at this code, the more $x is a plain old scalar and the more this condition will never be true. I'm calling it a bug at this point.
[hippo]: Thanks for your input which has soothed my sanity (a little)
[Corion]: Eily: Sure - if you force both things into stringy things, then you break that magic. But that would also mean that you changed the expression, as now $x = 0.00 will be true instead of false as it were before

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