Inspired by the upcoming winter bash event


Add a hat ^,´ or ` to a single vowel in each word of the input.


  • The hat and the vowel must be chosen at random. Each hat has to appear with the same probability (33%) and the vowels must have the same probability within the valid vowels in the word (if the word have 2 valid vowels each must be 50% likely to be chosen) - or the closest that your language has.
  • Only AEIOUaeiouare considered vowels (sorry y)
  • Vowels with hats in the input DO NOT interfere with the rules (you can consider it as a consonant)
  • If the input has no vowel it won't be modified
  • Capitalization must be preserved.


winter > wintér
bash > bâsh
rhythm > rhythm
rng ftw > rng ftw
cat in the hat > cât ìn thê hát
dès > dès
tschüss > tschüss
principî > prìncipî


This is so the shortest code wins

  • \$\begingroup\$ "at random (or the closest that your language has)" so if my language has no random commands I can just do something deterministic??? \$\endgroup\$
    – Fatalize
    Dec 16 '16 at 22:52
  • \$\begingroup\$ @LuisMendo better? \$\endgroup\$
    – Rod
    Dec 16 '16 at 22:55
  • \$\begingroup\$ @Fatalize not even get the current time random? \$\endgroup\$
    – Rod
    Dec 16 '16 at 22:57
  • \$\begingroup\$ @Rod I don't have any language in mind. I'm just saying that you have to impose that it is random (and what kind of random) otherwise it makes no sense at all. \$\endgroup\$
    – Fatalize
    Dec 16 '16 at 23:03
  • \$\begingroup\$ The specs say we have to add ^ ´ or backtick, but the example tschuss shows an umlaut. Are umlauts required or not? \$\endgroup\$ Dec 16 '16 at 23:18

C#, 273 267 bytes

using System.Linq;A=s=>{var r=new System.Random();var a=s.Split(' ');return string.Join(" ",a.Select(w=>w.Select((c,i)=>"AEIOUaeiou".Any(d=>c==d)?i:-1).Where(x=>x>=0).ToList()).Select((l,i)=>l.Any()?a[i].Insert(l[r.Next(l.Count)]+1,""+(char)r.Next(768,771)):a[i]));};

repl.it demo

I really feels like I'm cheating, since I still add hats to already accented vowels created by combining characters. If that is not acceptable, let me know so I can add boilerplate codes declare this answer non-competing.

This thing adds a random character among U+0300 or U+0301 or U+0302, after a random vowel of each input word (if any).

Ungolfed (lambda body only)

var r=new System.Random();
// Split sentence to array of words
var a=s.Split(' ');
// Join the (hat-ed) words back to sentence
return string.Join(
    " ",
    // Select an IEnum of list of integers indicating the positions of vowels
            // If it's vowel, return position (>=0), else return -1
        // Filter vowels only
    // Convert each list of integers to hat-ed words
            // Insert "something" into the word...
                // ...at the position after a random vowel in that word...
                // "something" is a random integer in [0x0300, 0x0303), then casted to UTF16 i.e. [U+0300, U+0302]
            // List is empty => just return original word

Perl 6, 101 bytes

~*.words.map: {(my$p=m:ex:i/<[aeiou]>/».to.pick)~~Mu:D??~S/.**{$p}<(/{("\x300".."\x302").pick}/!!$_}

Try it


~      # stringify (join with spaces)
*\     # Whatever lambda (input parameter)
.words # split into words

.map:  # for each word

{ # bare block lambda with implicit parameter 「$_」

    my $p =

    :exhaustive               # all possible matches
    /<[aeiou]>/\              # match a vowel

    ».to.pick                 # pick one of the positions

  ) ~~ Mu:D                   # is it defined ( shorter than 「.defined」 )

  ??                          # if 「$p」 is defined

    ~                         # stringify

      . ** {$p}               # match 「$p」 positions

      <(                      # ignore them
      ("\x300".."\x302").pick # pick one of the "hats" to add

  !!                          # if 「$p」 is not defined    
    $_                        # return the word unchanged

Mathematica, 226 bytes

Join@@((p=Position[w=#,Alternatives@@(v=Characters@"aeiouAEIOU")];If[p!={},i=#&@@RandomChoice@p;w[[i]]=FromCharacterCode[ToCharacterCode["àèìòùÀÈÌÒÙ"][[#&@@Position[v,w[[i]]]]]+RandomInteger@2]];w)&/@Split[#,{##}~FreeQ~" "&])&

Unnamed function taking a list of characters as input and returning a list of characters. Easier-to-read version, slightly ungolfed as well:

 1  v = Characters["aeiouAEIOU"];
 2  a = ToCharacterCode["àèìòùÀÈÌÒÙ"];
 3  Join @@ (
 4    (p = Position[w = #, Alternatives @@ v]; 
 5      If[p != {},
 6        i = First[RandomChoice[p]]; 
 7        w[[i]] =
 8          FromCharacterCode[
 9            a[[ First[ Position[ v, w[[i]] ] ] ]] + RandomInteger[2]
10          ]
11        ]; w
12    ) &
13  ) /@ Split[#1, FreeQ[{##1}, " "] &] &

Line 13 splits the input into words (sublists of characters) at all the spaces; each word is operated on by the function defined by lines 4-12, and the results joined back together again in a single list by line 3.

Line 4 sets p to the list of indices indicating which characters of the word w are vowels. If there are any vowels (line 5), we make a random choice of one such index i (line 6) and then reset that single character of the word to a new character (lines 7-10). Finally we output the (possibly modified) word w.

To select the new character, we find where the vowel to be replaced sits in the string v and choose the corresponding character code from a. But to randomly select the three hats, we take that code and add a random integer between 0 and 2 (line 9) before converting back to a character. (Fortunately the behatted vowels all come in consecutive trios of UTF-8 character codes.)


Python 3, 170 bytes

from random import *
print(' '.join([w,w[:i]+c('̀́̂')+w[i:]][i>0]for w in input().split()for i in[c([j+1 for j,x in enumerate(w)if x in 'aeiouAEIOU']or[0])]))


from random import choice
print(' '.join([
                   w,  # Don't modify the word if no vowels were found

                   w[:i] + choice('̀́̂') + w[i:]
               ][i > 0]
               for w in input().split()
                   for i in [choice([j + 1 for j, x in enumerate(w) if x in 'aeiouAEIOU']
                                    or [0])  # choice requires a non-empty sequence
  • 1
    \$\begingroup\$ You do not need the space between import and *. \$\endgroup\$
    – Mr. Xcoder
    Aug 29 '17 at 21:46
  • \$\begingroup\$ j+1 for can be j+1for. \$\endgroup\$ Feb 8 '18 at 15:09

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