In this challenge, you must take a string matching the regex ^[a-zA-Z]+$ or whatever is reasonable (you don't have to consider uppercase or lowercase letters if you want) (you may assume the string is long enough, and has the right structure for all the operations), and output another string, produced similarly to word at the end of a recent dadaist tweet by the POTUS ("Despite the constant negative press covfefe").

How to covfefify a string:

First, get the first sound group (made up terminology).

How do you do this? Well:

  • Find the first vowel (y is also a vowel)

  • Find the first consonant after that

  • Remove the rest of the string


That is your first sound group.

Next step:

Get the last consonant of the sound group


and replace it with the voiced or voiceless version. To do this, find the letter in this table. Replace with the letter given (which may be the same letter)

b: p
c: g
d: t
f: v
g: k
h: h
j: j
k: g
l: l
m: m
n: n
p: b
q: q
r: r
s: z
t: d
v: f
w: w
x: x
z: s

so, we get


Then, take the next vowel after that consonant. You can assume that this consonant is not at the end of the string. Join these two together, then repeat it twice:


Concatenate this to the first sound group:


You're done: the string is covfefified, and you can now output it.

Test cases:

coverage: covfefe

example: exxaxa

programming: progkaka (the a is the first vowel after the g, even though it is not immediately after)
code: codtete

president: preszizi

This is , so please make your program as short as possible!

  • 7
    \$\begingroup\$ "x" should technically map onto "gz". "qu" should map onto "gw". \$\endgroup\$ Jun 5 '17 at 5:34
  • 76
    \$\begingroup\$ Answers over 140 characters should be disqualified \$\endgroup\$ Jun 7 '17 at 15:27
  • 14
    \$\begingroup\$ Unfortunately it is impossible to do this in TrumpScript :( \$\endgroup\$
    – user69279
    Jun 14 '17 at 22:09
  • 5
    \$\begingroup\$ @ThePlasmaRailgun dude... \$\endgroup\$ Dec 20 '17 at 14:57
  • 9
    \$\begingroup\$ @ThePlasmaRailgun It was a joke, since tweets have to be 140 characters or less. \$\endgroup\$ Dec 31 '17 at 22:14

41 Answers 41


Jelly,  58  57 bytes


A full program that accepts a list of lowercase characters and prints the result.

Try it online!


<TḢị - Link 1, extract first value from y not less than x: number, x; list of numbers, y
     -                                                     e.g. 5, [3,4,7]
<    - x less than vectorised across y                             [0,0,1]
 T   - truthy indices                                              [    3]
  Ḣ  - head                                                             3
   ị - index into y                                                     7

e€Øyµ¬TĖEÐḟḢṪ;ç¥T - Link 2, indices of the letters to manipulate: list of characters, w
  Øy              - vowel+ yield = "AEIOUYaeiouy"                 e.g.  "smouching" 
e€                - exists in for €ach letter in w                       001100100
    µ             - monadic chain separation, call that v
     ¬            - not vectorised across v                              110011011
      T           - truthy indices                                       12  56 89
       Ė          - enumerate                      [[1,1],[2,2],[3,5],[4,6],[5,8],[6,9]]
         Ðḟ       - filter discard if:
        E         -   elements are equal                       [[3,5],[4,6],[5,8],[6,9]]
           Ḣ      - head                                        [3,5]
            Ṫ     - tail                                           5
                T - truthy indices of v                                    34  7
               ¥  - last 2 links as a dyad
              ç   -   call last link (1) as a dyad                         7
             ;    -   concatenate                                     5,7
                  -                                    ...i.e the indexes of 'c' and 'i'

ḣÇḢ⁸ÇịµḢØYiị“ßȷ%Hẹrȧq’œ?ØY¤⁾cgy;ẋ2 - Main link: list of characters, w
                                   -                             e.g.  "smouching"
 Ç                                 - call the last link (2) as a monad    [5,7]
ḣ                                  - head to index (vectorises)      ["smouc","smouchi"]
  Ḣ                                - head                             "smouc"
                                   -   implicit print due to below leading constant chain
   ⁸                               - link's left argument, w
    Ç                              - call the last link (2) as a monad    [5,7]
     ị                             - index into w                         "ci"
      µ                            - monadic chain separation, call that p
       Ḣ                           - head p                               'c'
        ØY                         - consonant- yield = "BCDFGHJKLMNPQRSTVWXZbcdfghjklmnpqrstvwxz"
          i                        - first index                          22
                          ¤        - nilad followed by link(s) as a nilad:
            “ßȷ%Hẹrȧq’             -   base 250 number = 1349402632272870364
                        ØY         -   consonant- yield = "BCDFGHJKLMNPQRSTVWXZbcdfghjklmnpqrstvwxz"
                      œ?           -   nth permutation  = "BCDFGHJKLMNPQRSTVWXZpctvkhjglmnbqrzdfwxs"
           ị                       - index into         (special case ->) 'c'
                           ⁾cg     - literal ['c','g']
                              y    - translate (change 'c's to 'g's)      'g'
                               ;   - concatenate with the headed p        "gi"
                                ẋ2 - repeat list twice                    "gigi"
                                   - implicit print ...along with earlier = smoucgigi
  • 14
    \$\begingroup\$ This is amazing... \$\endgroup\$
    – Klangen
    May 31 '17 at 14:23
  • \$\begingroup\$ Incredible work. \$\endgroup\$
    – JF it
    Jun 2 '17 at 13:22
  • 10
    \$\begingroup\$ I'm jelly. Upvoted. \$\endgroup\$
    – DeepS1X
    Jun 5 '17 at 5:01
  • 7
    \$\begingroup\$ This is the weirdest programming language I've ever seen. \$\endgroup\$
    – Ryan
    Jun 7 '17 at 16:57
  • \$\begingroup\$ @Ryan it's intended for golfing. \$\endgroup\$ Jul 17 '17 at 22:39

JavaScript (ES6), 107 103 bytes

Saved 4 bytes thanks to GOTO 0


Test cases

let f =



  • 6
    \$\begingroup\$ You can save a few bytes like this: s=>([,a,b,c]=s.match`(.*?[aeiouy]+(.)).*?([aeiouy])`,a+(b=(a="bcdfgszkvtgp")[11-a.search(b)]||b)+c+b+c) \$\endgroup\$
    – GOTO 0
    Jun 1 '17 at 0:13
  • \$\begingroup\$ @GOTO0 Thanks, updated. \$\endgroup\$
    – Arnauld
    Jun 1 '17 at 5:11

Jelly, 45 39 bytes


Try it online!

How it works

e€ØyIi-‘ɓḣ;ç                Main link. Argument: s (string)

  Øy                        Vowels with y; yield "AEIOUYaeiouy".
e€                          Test each character in s for membership.
    I                       Increments; compute the forward differences of the
                            resulting array of Booleans.
     i-                     Find the first index of -1.
       ‘                    Increment this index to find the index of the first
                            consonant that follows a vowel.
                            Let's call this index j.
        ɓ                   Begin a new chain. Left argument: s. Right argument: j
         ḣ                  Head; yield the first j characters of s.
           ç                Call the helper link with arguments s and j.
          ;                 Concatenate the results to both sides.
Øa“œṣ$b|0Ḃ’ṃ,Ṛ$yṫµfØyḢṭḢẋ2  Helper link. Left argument: s. Right argument: j

Øa                          Alphabet; set the return value to “abc...xyz”.
  “œṣ$b|0Ḃ’                 Yield 7787255460949942. This is a numeric literal in
                            bijective base 250. The value of each digit matches its
                            1-based index in Jelly's code page.
           ṃ                Convert 7787255460949942 to base 26, using the digts
                            a = 1, b = 2, ..., y = 25, z = 0.
                            This yields "bcdfkszgvtgp".
            ,Ṛ$             Pair the result with its reverse, yielding
                            ["bcdfkszgvtgp", "pgtvgzskfdcb"].
                ṫ           Call tail with arguments s and j, yielding the j-th and
                            all following characters of s.
               y            Translate the result to the right according to the
                            mapping to the left, i.e., replace 'b' with 'p', 'c'
                            with 'g', etc. 'g' appears twice in the first string
                            of the mapping; only the first occurrence counts.
                            Let's call the resulting string r.
                 µ          Begin a new chain. Argument: r
                  fØy       Filter; remove non-vowels from r.
                     Ḣ      Head; take the first vowel.
                       Ḣ    Head; take the first character/consonant of r.
                      ṭ     Tack; append vowel to the consonant.
                        ẋ2  Repeat the resulting string twice.
  • 4
    \$\begingroup\$ sorry, buddy, looks like you missed out on the mega jelly rep \$\endgroup\$ Jun 15 '17 at 12:17
  • \$\begingroup\$ tfw an answer looks overly simplistic but is in fact really wonderful...simple is beautiful \$\endgroup\$ Jun 16 '17 at 9:29

CJam, 59 58 57 56 bytes


Try it online!


q_                   e# Read the input and copy it.
{                    e# Find the index of the first char for which the following is true:
 "aeiouy":V          e#  Push "aeiouy" and store it in V.
 &,                  e#  Check if the current char is in the vowel string (0 or 1).
 _T|:T               e#  Copy the result and OR with T (T is initially 0), storing back in T.
 ^                   e#  XOR with the original result. This will be 1 for the first 
                     e#  consonant appearing after a vowel.
}#                   e# (end find)
)/                   e# Increment the index and split the string into chunks of that size.
(                    e# Pull out the first chunk.
_W>                  e# Copy it and get the last character (the consonant).
"cbdfkszgvtpg"_W%er  e# Transliterate the consonant to voiced/voiceless alternative.
@s                   e# Bring all the other split chunks to the top and join them together.
V&0=                 e# First char of the set intersection of that and the vowels.
                     e# (i.e. the first vowel in the second half)
+                    e# Concatenate the new consonant and the vowel.
_                    e# Duplicate the result of that.
                     e# Implicit output of stack contents.
  • 2
    \$\begingroup\$ CJam beats Jelly? :O (At least, it beats the Jelly answer everybody seems to be upvoting.) \$\endgroup\$ Jul 17 '17 at 22:41

C, 219 213 206 179 175 bytes

#define p putchar
#define q(a)for(;a strchr("aeiouy",*s);p(*s++));
f(s,c,h)char*s;{q(!)q()p(*s);p(c="pgt vkh jglmn bqrzd fwx s"[*s-98]);p(h=s[strcspn(s,"aeiouy")]);p(c);p(h);}

Try it online!

  • \$\begingroup\$ Does *p=putchar work as the first line? \$\endgroup\$
    – k_g
    Jun 3 '17 at 23:25
  • 6
    \$\begingroup\$ Sorry disqualified. Can't fit in a tweet. \$\endgroup\$ Jun 9 '17 at 6:46
  • 1
    \$\begingroup\$ 171 bytes \$\endgroup\$
    – ceilingcat
    Nov 30 '18 at 19:21
  • 1
    \$\begingroup\$ 12 or so bytes can be shaved off by replacing #defines and the function with preprocessor flags (-D...). \$\endgroup\$
    – user77406
    Dec 1 '18 at 15:53
  • 1
    \$\begingroup\$ Actually tweetable now because twitter allows 280 chars. \$\endgroup\$
    – lyxal
    Jan 22 '20 at 7:32

Perl 5, 81 72 bytes


Try it online!

  • \$\begingroup\$ You and I had the same idea with \K, but you did it 9 bytes better than I did. Good answer! \$\endgroup\$ Jun 6 '17 at 23:25

PHP, 121 Bytes


Try it online!

  • 3
    \$\begingroup\$ -2 bytes: echo$t[1],$z=strtr($t[2].$t[3],bcdfgkpstvz,pgtvkgbzdfs),$z; \$\endgroup\$
    – Titus
    Jun 1 '17 at 11:53
  • \$\begingroup\$ @Titus I have not think about that. Thank You \$\endgroup\$ Jun 1 '17 at 11:59
  • \$\begingroup\$ why not rename $argn to something shorter? $a, for example - that's -3 bytes \$\endgroup\$ Jun 6 '17 at 22:37
  • \$\begingroup\$ @TylerSebastian I must have an input variable that exists. Yes I can create a function but if I do it it raise the byte count more as use the three bytes \$\endgroup\$ Jun 6 '17 at 22:44
  • \$\begingroup\$ ah ok sorry I forgot how PHP does command line args - I just saw that you had defined it in the header section but failed to realize it was a reserved variable. \$\endgroup\$ Jun 6 '17 at 22:45

Pyth, 54 bytes


This defines a function y, that expects a string. Try it online: Test Suite


Python 3, 155 139 bytes

import re
def f(x,k='aeiouy])'):b,c,v=re.findall(f'(.*?[{k}([^{k}.*?([{k}',x)[0];return b+c+(('bcdfgkpstvz'+c)['pgtvkgbzdfs'.find(c)]+v)*2

removed 16 bytes thanks to @ovs

removed 1 byte thanks to Gábor Fekete

  • 2
    \$\begingroup\$ You could create a variable which has the value 'aeiouy]', maybe that will save some bytes. Also you can remove some characters from the replacement strings as there are the same. \$\endgroup\$ May 31 '17 at 12:41
  • 2
    \$\begingroup\$ I can't remove the identical characters from the replacement string, because that would be an IndexError, and saving aeiouy]) doesn't save any bytes. \$\endgroup\$
    – L3viathan
    May 31 '17 at 13:14
  • 2
    \$\begingroup\$ if you pull out something like s='aeiouy])', you could use b,c,v=re.findall('(.*?[%s([^%s.*?([%s'%(s,s,s). It's not shorter, but might lead towards a way to shorten it overall. \$\endgroup\$ May 31 '17 at 19:50
  • 6
    \$\begingroup\$ Shortened to 139 bytes \$\endgroup\$
    – ovs
    May 31 '17 at 20:34
  • 3
    \$\begingroup\$ Using f-strings will save 1 byte: k='aeiouy])' and f'(.*?[{k}([^{k}.*?([{k}' \$\endgroup\$ Jun 1 '17 at 7:48

Java 8, 243 236 222 bytes

s->{String q="[a-z&&[^aeiouy]]",a=s.replaceAll("(^"+q+"*[aeiouy]+"+q+").*","$1"),b="pgtvkhjglmnbqrzdfwxs".charAt("bcdfghjklmnpqrstvwxz".indexOf(a.charAt(a.length()-1)))+s.replaceAll(a+q+"*([aeiouy]).*","$1");return a+b+b;}

Uses .replaceAll regexes with capture groups to filter out the parts we don't want.


Try it here.

s->{ // Method with String parameter and String return-type
  // Temp String we use multiple times:
  String q="[a-z&&[^aeiouy]]",
   // Regex to get the first part (i.e. `creation` -> `creat` / `example` -> `ex`)
   // Get the trailing consonant and convert it
   // Get the next vowel after the previous consonant from the input-String
  // Return the result:
  return a+b+b;
} // End of method

Haskell, 143 141 138 137 136 bytes

z h=elem h"aeiouy"
f i|(s,(m,c:x))<-span z<$>break z i,j:_<-filter z x,d<-"pgt.vkh.jglmn.bqrzd.fwx.s"!!(fromEnum c-98)=s++m++[c,d,j,d,j]

Try it online!

  • 1
    \$\begingroup\$ Awesome! Replacing nx with something one-lettered will save 2 bytes. \$\endgroup\$
    – tomsmeding
    Jun 1 '17 at 7:08
  • \$\begingroup\$ declaring z outside of f and switching to guards instead of a let saves another two bytes: Try it online! \$\endgroup\$
    – Laikoni
    Jun 1 '17 at 7:52
  • 2
    \$\begingroup\$ And two more by combining (s,v)<-break z i,(m,c:x)<-span z v into (s,(m,c:x))<-span z<$>break z i. \$\endgroup\$
    – Laikoni
    Jun 1 '17 at 7:55
  • \$\begingroup\$ Could shave one more by putting the opening parenthesis next to the let, thanks! \$\endgroup\$
    – bartavelle
    Jun 1 '17 at 8:15
  • \$\begingroup\$ @Laikoni I don't understand the part about moving z out of f ? \$\endgroup\$
    – bartavelle
    Jun 1 '17 at 8:18

Python, 261 260 bytes

def c(s,t='bpcgdtfvgksz'):
 c=[i for i in r if i>[j for j in r if s[j]in q][0]and s[i]not in q][0]
 C=([t[2*i+1]for i in range(12)if s[c]==t[i*2]]or s[c])[0]
 return s[:c+1]+(C+s[[i for i in r if i>c and s[i]in q][0]])*2

A Non regex, Not esoteric solution. Took about 20 minutes to make, and an hour more to golf.

It probably has more list comprehension than the entire python standard library, mostly because I don't know regex...

Try it online! (With testcases)


Python 2, 251 246 245 239 237 234 229 211 bytes

First submission here.

def f(s):
  for i in s:
    if n<2:r+=i
    if n<1and i in w:n=1
    if n==1and i in a:c='pgtvkhjglmnbqrzdfwxs'[a.index(i)];n=2
    if n==2and i in w:r+=c+i+c+i;break
  return r

Try it online!

Fellow golfers that helped me:

 Destructible Lemon / Wheat Wizard - 5 bytes
 Hubert Grzeskowiak - 1 byte
 musicman523 - 16 bytes
  • 2
    \$\begingroup\$ Welcome to the site! I see you tried using tabs for indentation. If you replace each tab with a single space, it is functionally identical and actually shows up properly instead of as extra bytes \$\endgroup\$ Jun 1 '17 at 10:43
  • 4
    \$\begingroup\$ While what Destructible Lemon said is correct, you can save even more bytes in you source by indenting the first level of your code with a single space and the second level with a single tab, this will make it a bit hard to display, but will save you 5 bytes. \$\endgroup\$
    – Grain Ghost
    Jun 1 '17 at 10:45
  • 1
    \$\begingroup\$ Is the semicolon at the end of line 4 necessary? \$\endgroup\$ Jun 3 '17 at 17:54
  • 1
    \$\begingroup\$ @WaitndSee I think you can shorten some of your conditionals. First: you can change not n to n<1 for 2 bytes, since you know n will never be negative. Also you can change n==3 to n>2 since you know n will never be greater than 3. You can also use the Python tricks for conditionals to shorten the first and second-last even further: n=[n,1][i in w and n<1]; r+=[0,r][n<2] \$\endgroup\$ Jun 8 '17 at 4:33
  • 2
    \$\begingroup\$ You can change r,v,c=('',)*3 to r=v=c='', since strings are immutable. I've tried a bunch of other clever tricks but frustratingly they are exactly as long. Also it may be worth adding a Try it online! link to your post \$\endgroup\$ Jun 8 '17 at 13:14

Ruby, 90 bytes


Try it online!

Ungolfing it a bit, we have something equivalent to:

def covfefefify(x)
  v = '[aeiouy]'
  # Match x to a regular expression capturing:
  # Group 1:
  #  some characters (non-greedy)
  #  followed by some (greedy) non-zero number of vowels
  #  followed by exactly one character
  # Ungrouped:
  #  Some more (non-greedy) characters
  # Group 2
  #  Exactly one other vowel
  # By switching between greedy and non-greedy matches, we can capture longest and shortest vowel/consonant sequences without writing out all the consonants
  # Glue it back together, replace the necessary consonants, duplicate where needed

Ruby, 175 141 110 bytes


Try it online!

  • Saved 34 bytes thanks to Eric Duminil
  • Saved 31 bytes thanks to Value Ink + optimized suggested tr arguments


covfefify = -> (s) {
    from = 'bcdfgkpstvz'
    to   = 'pgtvkgbzdfs'

    vowels = "[aeiouy]"
    consonants = "[^aeiouy]"

    d = ($2.tr(from, to) + $3) * 2
  • 4
    \$\begingroup\$ -34 bytes with Hash[*"bpcgdtfvgkkgpbsztdvfzs".chars] \$\endgroup\$ May 31 '17 at 13:30
  • 1
    \$\begingroup\$ Since input seems guaranteed to be all alphabetical characters, c=[^aeiou] is shorter. Have the first interpolation for each variable assign it simultaneously for -2 bytes: /^(.*?${v='[aeiou]'}+(#{c='[^aeiou]})).../. Finally, $2.tr("b-z","pgtevkhijgl-obqrzdufwxys") instead of the Hash solution. \$\endgroup\$
    – Value Ink
    May 31 '17 at 18:42
  • \$\begingroup\$ You can save 14 bytes by using subexpressions (\g<n>) instead of interpolation, plus another 14 using @ValueInk's [^aeiou] suggestion: s=~/^(.*?([aeiouy])+([^aeiou]))\g<3>*(\g<2>)/. \$\endgroup\$
    – Jordan
    May 31 '17 at 20:02
  • \$\begingroup\$ Actually, that has a bug with programming -> progkaka, which I can't quite figure out. \$\endgroup\$
    – Jordan
    May 31 '17 at 20:19
  • \$\begingroup\$ @Jordan unfortunately the subexpression call \g<3> updates the value of $3, so we can't use this shortcut. \$\endgroup\$
    – sudee
    Jun 1 '17 at 9:15

Crystal, 203 194 187 186 184 163 bytes

o+=a if c<2||x
c>0&&(x&&break||(o+=(i="pgtvkgbqrzdfs"=~r)?"bcdfgkpqrstvz"[i]: a))}
p o+o[-2..-1]
  • \$\begingroup\$ I think you can lose the parens around c=v and o+=<...> \$\endgroup\$
    – Cyoce
    Jun 19 '17 at 15:24

MATLAB / Octave - 159 158 bytes

The following works assuming the input string is all lowercase.

a=input('','s');m=ismember(a,'aeiouy');s='pgt vkh jglmn bqrzd fwx s';m(1:find(m,1))=1;i=find(~m,1);f=a(1:i);d=s(f(end)-97);m(1:i)=0;v=a(find(m,1));[f d v d v]


  1. a = input('','s');: Gets a string from STDIN and stores it into the variable a.
  2. m=ismember(a,'aeiouy');: Returns a Boolean array that is the same size as the string a determining where vowels are located
  3. s='pgt vkh jglmn bqrzd fwx s'; The covfefe mapping of consonants as a string. This string is 25 characters long and omitting the vowels. The first position where the vowel 'a' is supposed to be is removed while the other positions where the vowels are located are placed with a dummy space character. This is so that when we determine the first consonant appearing after the vowel, we will convert the consonant to a position to access a character in this string to determine the first component of the converted word.
  4. m(1:find(m,1))=1: Sets the first position of the Boolean array up to where we have found the first vowel as all vowels. This will be so that when we search for the next consonant that follows the first vowel, we will ignore these characters.
  5. i=find(~m,1);: Finds the first position of the string that is a consonant after the first vowel.
  6. f=a(1:i): Removes the string after the first consonant that follows the vowel. We simply sample from the first position of the string up to this point.
  7. d=s(f(end)-97);: Take the last character of the string that is remaining and finds where we need to sample from the lookup string and gets that character. Subtracting a character and a number in MATLAB or Octave coalesces to form an integer by converting the character into its ASCII code. In this case, we subtract the last character by the character at the beginning of the alphabet to give us the position relative to the beginning. However, instead of subtracting by b (98), we subtract by a as MATLAB starts indexing by 1 instead of 0. 'a''s ASCII code is 97.
  8. m(1:i)=0;: Takes the Boolean mask and sets all characters in the input string from the first position to the first consonant following a vowel to false.
  9. v=a(find(m,1));: Finds the next vowel that follows the first consonant from the input string.
  10. [f d v d v]: Output our covfefeied string.

Example Runs

>> a=input('','s');m=ismember(a,'aeiouy');s='pgt vkh jglmn bqrzd fwx s';m(1:find(m,1))=1;i=find(~m,1);f=a(1:i);d=s(f(end)-97);m(1:i)=0;v=a(find(m,1));[f d v d v]

ans =


>> a=input('','s');m=ismember(a,'aeiouy');s='pgt vkh jglmn bqrzd fwx s';m(1:find(m,1))=1;i=find(~m,1);f=a(1:i);d=s(f(end)-97);m(1:i)=0;v=a(find(m,1));[f d v d v]

ans =


>> a=input('','s');m=ismember(a,'aeiouy');s='pgt vkh jglmn bqrzd fwx s';m(1:find(m,1))=1;i=find(~m,1);f=a(1:i);d=s(f(end)-97);m(1:i)=0;v=a(find(m,1));[f d v d v]

ans =


>> a=input('','s');m=ismember(a,'aeiouy');s='pgt vkh jglmn bqrzd fwx s';m(1:find(m,1))=1;i=find(~m,1);f=a(1:i);d=s(f(end)-97);m(1:i)=0;v=a(find(m,1));[f d v d v]

ans =


>> a=input('','s');m=ismember(a,'aeiouy');s='pgt vkh jglmn bqrzd fwx s';m(1:find(m,1))=1;i=find(~m,1);f=a(1:i);d=s(f(end)-97);m(1:i)=0;v=a(find(m,1));[f d v d v]

ans =


Try it online!


When you hit the Execute button at the top, wait a few moments, then enter the desired string. Enter the string slowly as there seems to be a delay when entering in text.


Clojure, 182 156 chars

#(let[v #{\a\e\i\o\u\y}p(partition-by v %)[s m[c][n]](if(v(first %))(cons[]p)p)z[(or((zipmap"bcdfgkpstvz""pgtvkgbzdfs")c)c)n]](apply str(concat s m[c]z z)))

How It Works

(partition-by v "president")

Returns a seq of ((\p \r) (\e) (\s) (\i) (\d) (\e) (\n \t))

[s m [c] [n]] (if (v (first x)) (cons [] p) p)

Destructures the seq into s=(\p \r), m=(\e), c=\s, n=\i.

Or for "example" it's s=[], m=(\e), c=\x, n=\a.

(apply str (concat s m [c] [(l c) n] [(l c) n]))

Returns the output string by concatenating the pieces together and stringifying it.

And then I just removed as much whitespace as I could while still making it compile.


(defn covfefify [x]
  (let [vowel? #{\a\e\i\o\u\y}
        parts (partition-by vowel? x)
        [start mid [consonant] [last-vowel]] (if (vowel? (first x)) (cons [] parts) parts)
        lookup #(or ((zipmap "bcdfgkpstvz" "pgtvkgbzdfs") %) %)]
    (apply str (concat start mid [consonant] [(lookup consonant) last-vowel] [(lookup consonant) last-vowel]))))
  • \$\begingroup\$ Welcome to PPCG, and great first answer! We hope you'll stay and have fun participating in more challenges. :-) \$\endgroup\$ Jun 6 '17 at 22:23
  • \$\begingroup\$ If you're defining a function, its name should probably be as short as possible. You could just call the main function c, for example. (We also allow anonymous functions, which are shorter in many languages; I'm not sure whether they are in Clojure). I see you've made that improvement in the interior of your code already, though, so probably not much needs changing here. \$\endgroup\$
    – user62131
    Jun 6 '17 at 22:31

R, 341 characters


Horrendous R attempt, why are strings so hard

Readable version:

f = function(x) {
  g = function(x, y)el(strsplit(x, y))
  a = g(x, '')
  v = g('aeiouy', '')
  n = letters[-c(1, 5, 9, 15, 21, 25)]
  l = data.frame(n, g('pgtvkhjglmnbqrzdfwxs', ''))
  y = min(match(n, a)[which(match(n, a) > min(match(v, a), na.rm = T))])
  m = l[which(l$n == a[y]), 2]
  e <-a[-c(1:y)][min(match(v, a[-c(1:y)]), na.rm = T)]
  paste0(paste0(a[c(1:y)], collapse = ''), m, e, m, e, collapse = "")

Python 3.8 (pre-release), 142 bytes

g=lambda i,f='aeiuoy':i if s[i]in f else g(i+1,f)
q=g(g(0),c:='pgtcvkh jglmn bqrzd fwx s')

Try it online!

A little late to the party, but here's yet another non-regex Python answer! I interpreted the rules to allow printing to STDERR which saves a byte (exit/print). Using Python 3.8 over 3<=3.7 saves me a total of 1 byte with the walrus operator as opposed to defining the c variable elsewhere.

Thanks a lot to Post Rock Garf Hunter (-21 bytes) for the help!

  • 2
    \$\begingroup\$ Looks good! +1 for the good job :D also, you are never late for a code-golf party! \$\endgroup\$
    – RGS
    Feb 13 '20 at 14:08
  • 1
    \$\begingroup\$ You can take off some bytes by having b look for the index instead of the character since indexing is shorter than .find. Try it online! \$\endgroup\$
    – Grain Ghost
    Feb 13 '20 at 15:03
  • \$\begingroup\$ I really, really like this, but does it not break on input words such as peeled? Try it online! I need a bit of time to properly understand it, I can't quite figure out how to fix it but I have a feeling that it is possible! \$\endgroup\$
    – chinatsu
    Feb 13 '20 at 15:49
  • \$\begingroup\$ I thought I had it with this, however it also breaks if the word start with a vowel :( The example given should return avfefe \$\endgroup\$
    – chinatsu
    Feb 13 '20 at 16:31
  • 1
    \$\begingroup\$ You can save a few bytes with short-circuiting \$\endgroup\$
    – squid
    Feb 13 '20 at 16:51

Retina, 68 bytes


Try it online!


BlitzMax, 190 bytes

s$=Input()For i=1To s.Length
f="aeiouy".Contains(s[i-1..i])If f v=i If c Exit
If v And c|f=0c=i
t$="bpdtfvgkcgsz"x$=s[c-1..c]r=t.Find(x)~1If r>=0x=t[r..r+1]
x:+s[v-1..v]Print s[..c]+x+x

Takes a word from stdin and prints the result to stdout. The input word is assumed to be lowercase and to contain at least one vowel followed by a consonant.

A more readable version of the progam with formatting and variable declarations:

Framework BRL.StandardIO

Local s:String = Input()
Local v:Int
Local c:Int

For Local i:Int = 1 To s.Length
    Local f:Int = "aeiouy".Contains(s[i - 1..i])
    If f Then
        v = i
        If c Then Exit
    End If
    If v And c | f = 0 Then c = i

Local t:String = "bpdtfvgkcgsz"
Local x:String = s[c-1..c]
Local r:Int = t.Find(x) ~ 1
If r >= 0 Then x = t[r..r + 1]
x :+ s[v - 1..v]
Print s[..c] + x + x

How it works:

BlitzMax doesn't have any builtin regex functionality or similar, so a loop is used to iterate over the characters of the input word until it finds a vowel followed by a chain of at least one consonant. The variable c stores the position of the last of those consonants, v that of the vowel. The loop continues to see if there is another vowel after the chain and if so, v is updated accordingly. Then the consonant at c is looked up in the string "bpdtfvgkcgsz", which acts as a replacement table. If the consonant is found in the table at any position, then that position is XOR-ed with 1 and the character at the resulting position gets used as its replacement. The XOR operation turns 0 into 1, 2 into 3, 4 into 5 etc. and vice versa, so that b gets swapped with p, d with t and so on. Finally, the original string up to c, the replacement character and the vowel at v are put together as required and printed.

Example results:

coverage covfefe

creation creatdidi

programming progkaka

stupidity stupbibi

blah blahhaha

  • \$\begingroup\$ link to blitzmax repo? \$\endgroup\$ Jun 3 '17 at 6:29
  • \$\begingroup\$ @DestructibleLemon BlitzMax was created as a language primarily for amateur game delevopment and with a proprietary compiler sold for money. While it is now free and available from here, I believe the compiler is still not open source. There exists an alternative implementation (repo here, builds here), which will however only run the ungolfed version of above code due to the lack of a "non-strict" setting that allows omitting variable declarations. \$\endgroup\$ Jun 3 '17 at 7:55
  • \$\begingroup\$ Clever use of XOR on index -- I'll probably be using that someday. Thank you. \$\endgroup\$ Jun 8 '17 at 3:42

Perl, 71 bytes


Also run with perl -pe. A few bytes less than the previous Perl solution. Admittedly I got some inspiration from there as well.


05AB1E, 101 104 88 bytes

-16 bytes thanks to Okx

I somehow hope this can be done way more efficiently.


Try it online!


                  Argument: s
žOÃ0èk            Get index of first vowel in s
>[DIs£¤žPså#\>]   Increment index and split s until last character of substring is a consonant
s[DIsèDžOså#\>]   Increment index an get character at index in s until character is a vowel
ŠŠ                Rearrange stack
.•7¶ëÒ—Öb´ƒ≠Ä“šʒƵJ^ÝV“Îpи•S2ôDí«ø`   Prepare character substitution map
Šs                Rearrange stack
¤                 Last character of substring
sŠ                Rearrange stack (yes, again)
k                 Index of last character in substitution key list
sŠ                Rearrange stack (it won't stop)
è                 Character at index in character substitution value list
sŠ                Rearrange stack (ONE LAST TIME)
ì2׫              Prepend substitution consonant before vowel, duplcicate and concatenate with the substring from the very beginning
  • \$\begingroup\$ You can replace "bpcgdtfvgkhhjjkgllmmnnpbqqrrsztdvfwwxxzs" with .•7¶ëÒ—Öb´ƒ≠Ä“šʒƵJ^ÝV“Îpи• to save 15 bytes \$\endgroup\$
    – Okx
    Jun 19 '17 at 9:41
  • \$\begingroup\$ You can also replace žOÃćIsk with žOÃ0èk to save another byte. \$\endgroup\$
    – Okx
    Jun 19 '17 at 9:47
  • \$\begingroup\$ @Okx I think I really need to learn some String compression techniques. Thanks! \$\endgroup\$
    – kalsowerus
    Jun 19 '17 at 12:41
  • \$\begingroup\$ @kalsowerus I know it's been a while, but you can golf 8 bytes from your answer like this: žOÃнk>[DIs£¤žPså#\>]©s[DIsèDžOså#\>]s\.•7¶ëÒ—Öb´ƒ≠Ä“šʒƵJ^ÝV“Îpи•S2ôDí«ø`®θkèìDJ Try it online. I mainly got rid of all the swaps and triple-swaps by using a variable instead. And can be н, and I've replaced 2׫ with DJ to join the entire stack together. PS: I've also posted a 55 bytes 05AB1E answer using a different technique. (Which also includes a link to better understand compression in 05AB1E. :D) \$\endgroup\$ Dec 1 '18 at 11:27

05AB1E, 55 42 bytes


-13 bytes thanks to @Grimmy.

Try it online or verify all test cases.


η                # Suffixes of the (implicit) input
                 #  i.e. "creation" → ["c","cr","cre","crea","creat","creati","creato","creatio","creation"]
 .Δ       }      # Find the first for which the following is truthy:
   žO            #  Push vowels (including y): "aeiouy"
     S           #  Convert it to a list of characters: ["a","e","i","o","u","y"]
      å          #  Check for each if they're in the current (implicit) suffix
                 #   i.e. "creat" → [1,1,0,0,0,0]
       à         #  Pop and push the max (basically check if any are truthy)
                 #   i.e. [1,1,0,0,0,0] → 1
   y             #  Push the suffix again
    žP           #  Push the consonants (excluding y): "bcdfghjklmnpqrstvwxz"
      S          #  Convert to a list of characters: ["b","c","d","f","g","h","j","k","l","m","n","p","q","r","s","t","v","w","x","z"]
       Å¿        #  Check for each if the suffix ends with it
                 #   i.e. "creat" → [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0]
         à       #  Pop and push the max (basically check if any are truthy)
                 #   i.e. [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0] → 1
   *             #  Check if both are truthy
                 #   i.e. 1 and 1 → 1
           Ð     # Triplicate the found suffix
            I    # Push the input
             s   # Swap the top two items on the stack
                 #  i.e. stack contains now: "creat","creat","creation","creat"
K                # Remove the suffix from the input
                 #  i.e. "creation" and "creat" → "ion"
 žOÃ             # Only leave the vowels
                 #  i.e. "ion" → "io"
    н            # Pop and push the first character
                 #  i.e. "io" → "i"
s                # Swap again so the prefix is a the top of the stack again
 θ               # Pop and push the last character
                 #  i.e. "creat" → "t"
  .•gÍĆdQ¸G•     # Push string "bcdfkszgvtgp"
            Â    # Bifurcate it (short for Duplicate & Reverse copy): "pgtvgzskfdcb"
             ‡   # Transliterate the character of "bcdfkszgvtgp" to the same index in "pgtvgzskfdcb"
              ì  # Prepend the second character in front of the first
                 #  i.e. "d" and "i" → "di"
               D # Duplicate it
J                # Join the stack together (and output implicitly)
                 #  i.e. "creat" and "di" and "di" → "creatdidi"

See this 05AB1E tips of mine (section How to compress strings not part of the dictionary?) to understand why .•gÍĆdQ¸G• is "bcdfkszgvtgp".

  • 1
    \$\begingroup\$ I ended up making my own answer, since even though I used your answer as a starting point, the result is very different. Unfortunately couldn’t get it below 35. \$\endgroup\$
    – Grimmy
    Feb 1 '20 at 12:44

Crystal, 130 Bytes

x,y,z=ARGV[0].partition /[^aeiouy]*#{c}*/
b=((i="pgtvkgbqrzdfs"=~/#{k}/)?"bcdfgkpqrstvz"[i]: k)+z[c]
p y+k+b*2

How it works

c = /[aeiouy]/

store a regex for searching first vowel to c.

x, y, z = ARGV[0].partition /[^aeiouy]*#{c}*/

split the first argument into three parts {"", String until one character before the first consonant after first vowel, rest of string} and store each of the elements into x, y and z.

k = z[0]

get the first character, the relevant consonant.

i = "pgtvkgbqrzdfs" =~ /#{k}/

get the index of the consonant inside the left string or nil.

b = ((i = ...) ? "bcdfgkpqrstvz"[i] : k) + z[c]

if i is not nil, use this index for the second string (kind of a golfed hash).

if i is nil, use the original character.

next, append the first vowel of z.

p y + k + (b * 2)

finally, print first part from first regex y, the first consonant k and two times the previous calculated string b.

Try it online.


SmileBASIC 3, 195 bytes

Very late to this question, but how could I resist a good challenge for SmileBASIC 3? Features like iterating over a sequence or manipulating a string aren't quite as robust as other languages, so this is a bit of a challenge to do it as small as possible. Assumes words are UPPERCASE.


Detailed explanation here!


05AB1E, 35 bytes


Try it online!

1ú                 # prepend the input with a space
                   # (this ensures the word doesn’t start with a vowel)
  .γ    }          # group characters by:
    žO             #  built-in constant "aeiouy"
      så           #  is the character in that string?
         R         # reverse the list of groups
          `        # dump all on the stack (first group on top)
¦                  # remove the space we added earlier
 ?                 # print the first consonant group
  ?                # print the first vowel group
   н               # get the first letter of the second consonant group
    ©              # save it in the register
     ?             # print it
      н            # get the first letter of the second vowel group
       ®           # restore the consonant from the register
        ì          # prepend, giving a consonant-vowel pair
.•gÍĆdQ¸G•         # compressed string "pgtvgzskfdcb"
          Â        # push a reversed copy of it
           ‡       # transliterate (b => p, c => g, ...)
            D      # duplicate the transliterated pair
             ?,    # print both copies (with a newline the second time)

Lua, 164 157 bytes


Edit 1: Removed 7 bytes by looking for any character after the consonants (see regex)

Try it online!

This program takes a string in CLI argument and prints its covfefied version.

This is my first submission to a code golf! I didn't check the others in detail so I might have missed some common optimizations (and fell in some traps). I used Lua because I've grown to like this little language, and I tried to find a regex that suited my needs.

Here's a cleaner version, using a function (I intended to use one, but the keywords in Lua are too long!):

function covfefy(word)
  i, j, a, b = word:find('[aeiouy]+([^aeiouy]+)(.)')

  -- 'a' is one or several consonants following the first vowel, b is the first vowel after that
  -- 'i' is the index of the beginning of 'a', 'j' the index of 'b'

  cov = word:sub(1, j - #a)

  -- Look for the first letter of 'a' in the voiced/voiceless table
  f = ('pgtvkhjglmnbqrzdfwxs'):sub(('bcdfghjklmnpqrstvwxz'):find(a:sub(1, 1)))

  return cov .. (f .. b):rep(2)

Feel free to give some feedback :)

Note: If you're wondering, it's 149 bytes long using MoonScript!


JavaScript (ES5), 237 229 bytes

function(s){r=['aeiouy','bcdfgkpstvz','pgtvkgbzdfs']i=0,p=''while(p+=s[i],r[0].indexOf(s[i++])<0);while(p+=s[i],~r[0].indexOf(s[i++]));b=s[i-1];while(r[0].indexOf(s[i++])<0);c=r[1].indexOf(b)d=((~c)?r[2][c]:b)+s[i-1]return p+d+d}

Try it online!

Probably not the most golfy, but it is ES5.

Recently fixed a bug. Example output:


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