There are two forms of nouns, singular and plural. The conversion between these two is quite easy.

  1. Normally, you end it with s. ex. car => cars.

  2. If it ends with s,x,z,ch or sh, end it with es. ex. bus=>buses.

  3. If it ends with y with a consonant just before it, change the y to ies. ex. penny => pennies.

  4. If it ends with f or fe, change it to ves. ex. knife => knives.

  5. If it ends with o with a consonant just before it, change it to oes. ex.potato => potatoes.


You will be given a singular noun. You have to convert the given noun to plural and output it.


  • You will not be given irregular nouns, like mouse and moose.

  • You will not be given exceptions, such as safe (safes; violating #4), piano (pianos; violating #5) and o (oes, violating #5).

  • You will not be given words which have two or more possible plural forms, such as mosquito (mosquitos or mosquitoes) and roof (roofs or rooves).

  • You will not be given uncountable nouns.

  • y doesn't count as a vowel.


car => cars
bus => buses
potato => potatoes
knife => knives
penny => pennies
exception => exceptions
wolf => wolves
eye => eyes
decoy => decoys
radio => radios
  • \$\begingroup\$ Edited question for clarity. Feel free to rollback. \$\endgroup\$ – JungHwan Min Mar 5 '17 at 17:03
  • 10
    \$\begingroup\$ Ahh, English - a huge pile of arbitrary rules and special cases :) \$\endgroup\$ – Esolanging Fruit Mar 5 '17 at 23:16
  • 36
    \$\begingroup\$ @Challenger5 Yep, but you can understand it through tough thorough thoughts, though. ;) \$\endgroup\$ – JungHwan Min Mar 5 '17 at 23:20
  • \$\begingroup\$ @MatthewRoh I've edited the consonant in front rule to make it clearer. Also added a couple of test cases for the same. If I've misunderstood, please edit it to clarify. \$\endgroup\$ – ghosts_in_the_code Mar 6 '17 at 4:20
  • 2
    \$\begingroup\$ @Challenger5 If you compare English to Dutch there are barely any rules at all.. Dutch has rules and special cases, and special cases contradicting those special cases, and in some cases even special cases that contradict those special cases that those special cases contradict. ;) \$\endgroup\$ – Kevin Cruijssen Mar 6 '17 at 8:13

17 Answers 17


Mathematica, 9 bytes


Yes, there is a built-in for this!

Sample output



Pluralize /@ {"bus", "potato", "knife", "penny", "exception", "wolf", "eye"}

{"buses", "potatoes", "knives", "pennies", "exceptions", "wolves", "eyes"}

  • 5
    \$\begingroup\$ Waaaaaat! Is there something Mathematica has no built-in for? \$\endgroup\$ – KeyWeeUsr Mar 6 '17 at 20:28
  • 1
    \$\begingroup\$ D: Builtins have attacked this challenge too \$\endgroup\$ – Matthew Roh Mar 15 '17 at 13:18
  • \$\begingroup\$ @KeyWeeUsr codegolf.stackexchange.com/a/71680/56033 \$\endgroup\$ – Azor Ahai Aug 25 '17 at 23:42

Retina, 57 53 56 55 58 57 bytes

Thanks to MartinEnder for some golfing suggestions

Thanks to BusinessCat for golfing 1 byte


Try it online!

Explanation (outdated)


Changes {consonant}y to {consonant}ie


Appends an e to when the word ends with an {consonant}o, f,x,z,s,sh or ch.


Changes an ending fe to ve


Finally append an s to the word.


  • Added bytes because I forgot the second rule
  • Added bytes to update with eye as an example
  • 1
    \$\begingroup\$ Sorry if this is a stupid question, I've not used Retina. Why are the round brackets needed in the first line? \$\endgroup\$ – user2390246 Mar 6 '17 at 7:21
  • \$\begingroup\$ Never mind, I think I've answered my own question. It's because of the lookback reference in the following line. \$\endgroup\$ – user2390246 Mar 6 '17 at 8:45
  • \$\begingroup\$ Yeah, it's because we want to capture the character before the y using $1 \$\endgroup\$ – Kritixi Lithos Mar 6 '17 at 11:00
  • \$\begingroup\$ I think I got it in 57 bytes: Try it online \$\endgroup\$ – Business Cat Mar 6 '17 at 15:56

JavaScript (ES6), 109 bytes


Test cases

let f =


console.log(f("car"));        // => cars
console.log(f("bus"));        // => buses
console.log(f("potato"));     // => potatoes
console.log(f("knife"));      // => knives
console.log(f("penny"));      // => pennies
console.log(f("exception"));  // => exceptions
console.log(f("wolf"));       // => wolves
console.log(f("eye"));        // => eyes

  • \$\begingroup\$ 'iv'[i]||'' could be replaced with the shorter 'iv\0'[i]. But is it OK to insert a NUL character? \$\endgroup\$ – Arnauld Mar 5 '17 at 17:55
  • \$\begingroup\$ Why do you have a () in front of fe? \$\endgroup\$ – Kodos Johnson Mar 5 '17 at 19:05
  • 1
    \$\begingroup\$ @KodosJohnson All replace() iterations include a reference to the first matching group (with $1). That's why I need an empty matching group here. \$\endgroup\$ – Arnauld Mar 5 '17 at 19:11
  • \$\begingroup\$ Have you tried (?<![aeiou])y? \$\endgroup\$ – Titus Mar 6 '17 at 9:25
  • \$\begingroup\$ @Titus Unfortunately, JS doesn't implement lookbehind assertions. \$\endgroup\$ – Arnauld Mar 6 '17 at 11:45

Batch, 325 bytes

@for %%v in (a e i o u)do @(
for %%e in (o y)do @if %s:~-2%==%%v%%e goto s
if %s:~-2%==%%vf set s=%s:~,-1%ve&goto s
if %s:~-3%==%%vfe set s=%s:~,-2%ve&goto s
@if %s:~-1%==y set s=%s:~,-1%ie
@for %%e in (o s x z)do @if %s:~-1%==%%e set s=%s%e
@for %%e in (c s)do @if %s:~-2%==%%eh set s=%s%e
@echo %s%s
  • \$\begingroup\$ What about @echo off at the beginning rather than @ everywhere? Also, @set/ps= seems a little bit rusty from a phone. Won't the s variable accept the slicing values anyway? \$\endgroup\$ – KeyWeeUsr Mar 9 '17 at 7:44
  • \$\begingroup\$ @KeyWeeUsr @echo off is already 9 bytes without the newline, so it doesn't save me anything. Also, @set/ps= is needed to input the value in the first place. \$\endgroup\$ – Neil Mar 9 '17 at 8:46

Haskell, 216 207 205 bytes

Thanks to @Lynn, @user1472751 and @Laikoni for the help!

import Data.List
p s|s!(words"s x z ch sh"++map(:"o")c)=s++"es"|s!map(:"y")c=init s++"ies"|s!["f"]=init s++"ves"|s!["fe"]=(init.init)s++"ves"|0<1=s++"s"


import Data.List;

endsWithOneOf :: String -> [String] -> Bool
endsWithOneOf str ends = (or . map (\end -> end `isSuffixOf` str)) ends 

consonants :: [Char]
consonants = ['a'..'z'] \\ "aeiou"

pluralize :: String -> String
pluralize str
    | str `endsWithOneOf` (words "s x z ch sh" ++ (map (:"o") consonants)) = str ++ "es"
    | str `endsWithOneOf` (map (:"y") consonants) = init str ++ "ies"
    | str `endsWithOneOf` ["f"] = init str ++ "ves"
    | str `endsWithOneOf` ["fe"] = (init.init) str ++ "ves"
    | otherwise = str ++ "s"


import Data.List for the function isSuffixOf. endsWithOneOf ( in the golfed version) returns whether one of the list elements is an ending of the string. consonants(c) is just a list of all consonants.

Finally, pluralize(p) checks for the endings and returns the proper pluralization.


p "potato" == "potatoes"
  • 1
    \$\begingroup\$ Nice solution! This is 216 characters, but is multiple bytes long, making your solution 226 bytes. (Code golf challenges are explicitly scored in bytes, because counting characters lets you cheat sometimes.) You can just rename it to !, though! Also, words"s x z ch sh" saves 5 bytes. Removing parens around (map(:"o")c)) and (map(:"y")c)) saves 4 more. \$\endgroup\$ – Lynn Mar 6 '17 at 9:58
  • \$\begingroup\$ Thanks for the help, @Lynn! I implemented your suggestions. \$\endgroup\$ – Eisfunke Mar 6 '17 at 14:56
  • 2
    \$\begingroup\$ You can save one byte by using c=['b'..'z']\\"eiou" since 'a' is always removed. \$\endgroup\$ – user1472751 Mar 6 '17 at 15:59
  • 1
    \$\begingroup\$ 0<1 is one byte shorter than True. Also newlines are the same byte count as ; but make the golfed code a bit better readable. \$\endgroup\$ – Laikoni Mar 7 '17 at 19:13

Perl, 66 + 2 (-pl flag) = 68 bytes



perl -ple '$_.=/(ch|sh?|x|z|[^aeiou]o)$/+s/([^aeiou])y$/$1i/+s/fe?$/v/?es:"s"' <<< car

Try it on Ideone.


Röda, 80 bytes


The function modifies its argument. Usage: main word { f word; print word } Here's a version that uses a return value (83 bytes):

f s{s~="([^aeiou])y$","$1ie","([sxz]|[cs]h|[^aeiuo]o)$","$1e","fe?$","ve";[s.."s"]}

And below is a function that reads infinitely many values from the input stream and pushes plural forms to the output stream (87 83 bytes):


It's an anonymous function, as that is shorter than creating a named function.

  • \$\begingroup\$ How do you get to display the result of the first function (the one starting with f&s)? Simply f("word") doesn't seem to display anything \$\endgroup\$ – Kritixi Lithos Mar 5 '17 at 19:02
  • \$\begingroup\$ @KritixiLithos The parameter is a reference, so the argument must be a variable. \$\endgroup\$ – fergusq Mar 5 '17 at 19:04

PHP, 103 100 bytes


Try it online!

The preg_replace function takes in an array of patterns and replacements.

  • Saved 2 bytes thanks to Titus.
  • Saved 1 byte thanks to Dewi Morgan.
  • 2
    \$\begingroup\$ I think You can save one byte with -R and $argn. And using an assertion with y saves two: (?<![aeiou])y$ allows ie as replacement: no \1, no quotes. \$\endgroup\$ – Titus Mar 6 '17 at 9:17
  • 1
    \$\begingroup\$ Another byte from ([^aeiou]o|sh?|x|z|ch)$ \$\endgroup\$ – Dewi Morgan Mar 6 '17 at 17:54
  • \$\begingroup\$ @Titus Actually it looks like there is a 1 byte penalty for using -R (but not -r) so that doesn't change the byte count, unfortunately. But the lookbehind suggestion works great. Thanks. \$\endgroup\$ – Kodos Johnson Mar 6 '17 at 19:32

Python 3, 271 239 199 bytes

Thanks to @ovs for reducing it by 72 bytes!

lambda s,v="aeiou":(s[-2:]=="fe"and s[:-2]+"ve"or s[:-1]+((s[-1]=="y"and s[-2]not in v)*"ie"or s[-1]=="f"and"ve"or s[-1]+((s[-1]in"sxz"or s[-2:]in["ch","sh"])+(s[-1]=="o"and s[-2]not in v))*"e"))+"s"

Try it online!

  • 1
    \$\begingroup\$ You can remove some unnecessary whitespaces and combine the first and last elif. The single character lists can be replaced by strings. Switching to python saves additional 3 bytes. tio \$\endgroup\$ – ovs Mar 6 '17 at 6:37
  • \$\begingroup\$ @ovs Done, thanks! I didn't combine the elifs however, because that means potato becomes potaties. \$\endgroup\$ – numbermaniac Mar 6 '17 at 6:47
  • 1
    \$\begingroup\$ I looked in the wrong line ;). You can combine the if with the last elif. To save some more bytes replace the last line with print(s+"s") and remove the else case as well every s you are appending to the word. Tio \$\endgroup\$ – ovs Mar 6 '17 at 7:06
  • 1
    \$\begingroup\$ When you replace your if/elif logic with and/* and or/+ and make an unnamed lambda function you can get it under 200 bytes (I swapped the cases a little bit) \$\endgroup\$ – ovs Mar 6 '17 at 7:35
  • \$\begingroup\$ @ovs Ooh, that print(s+"s") is clever. All changed; you pretty much rewrote the whole thing lol. Thanks! (I didn't even know you could do True and "string" like that) \$\endgroup\$ – numbermaniac Mar 6 '17 at 7:51

sed, 70 79 bytes

69 78 + 1 for -E (BSD)/-r (GNU) flag


Direct port of the retina answer.


Pip, 63 61 bytes


So close to catching Retina! But it's probably not going to happen. :(

Try it online!


Basic strategy: Replace performs several replacements one after the other when given lists of patterns and replacements. We want to make the following replacements:

  • (sh?|x|z|ch)$ -> add an e
  • [^aeiou]y -> change the y to i and add an e
  • fe? -> change to v and add an e
  • [^aeiou]o -> add an e

Then we want to tack on an s regardless.


  • The C operator, given a regex, wraps it in a capturing group; C`xyz` is one byte shorter than `(xyz)`.
  • A list of regexes or replacements that all end with the same character can be created by concatenating the character to the list instead of including it in all the items. Concatenating a Scalar (string) to a Pattern (regex/replacement) coerces to a Pattern.
  • Instead of concatenating the s (and having to deal with the precedence ordering of R and .), we can simply Output the main part of the word and then print the s separately.

Spaced and commented code:

                  a is 1st cmdline input (implicit)
Y`[^aeiou]`       Yank the consonant regex into the y variable
O a R             Output (without newline): a, with the following replacements:
 [                List of regexes to replace:
  C `sh?|x|z|ch`    (sh?|x|z|ch)
  Cy . 'y           ([^aeiou])y
  `fe?`             fe?
  y . 'o            [^aeiou]o
 ] . '$           End of list; concatenate $ to each item
 [                List of replacements:
  _                 Identity function (replace with whole match)
  B                 B is short for {b}, a function returning its second argument; as a
                    callback function for regex replacement, the second argument is
                    the value of capturing group 1 (the consonant before y)
    . 'i            To that, concatenate i
  'v                Scalar literal v
  _                 Identity function
 ] . 'e           End of list; concatenate e to each item
's                Return Scalar literal s, which is autoprinted

C#, 73 163 bytes:


Yes, another language with it built-in (although you need to add a reference to System.Data.Entity.Design.dll)

To use:

var words = new[] { "car", "bus", "potato", "knife", "penny", "exception", "wolf", "eye", "decoy", "radio" };
foreach (var word in words)
    var plural = p(word);
    Console.Out.WriteLine($"{word} => {plural}");


car => cars
bus => buses
potato => potatoes
knife => knives
penny => pennies
exception => exceptions
wolf => wolves
eye => eyes
decoy => decoys
radio => radios
  • \$\begingroup\$ Welcome to the site. How do I run this code? \$\endgroup\$ – Wheat Wizard Mar 6 '17 at 16:06
  • \$\begingroup\$ @WheatWizard updated. Should I have included more detail (using statements etc) in the byte count? \$\endgroup\$ – RoadieRich Mar 6 '17 at 18:28
  • \$\begingroup\$ Interesting bit of trivia, the reverse of this (Singularize) fails quite a few simple test cases. For example, it's convinced the singular of "courses" is "cours". \$\endgroup\$ – Morgan Thrapp Mar 6 '17 at 21:42
  • \$\begingroup\$ I think the namespaces needs to be included in this one's byte count, especially given that it's not one of the 'normal' ones. But I think you also need to at least wrap this in a lambda, passing the argument to the method. As is this is just a method group \$\endgroup\$ – pinkfloydx33 Mar 7 '17 at 0:34
  • \$\begingroup\$ @pinkfloydx33 better now? \$\endgroup\$ – RoadieRich Mar 7 '17 at 16:18

Python 199 187 176 Bytes

lambda s:s+'\bve'*(s[-1]=='f')+'\b\bve'*(s[-2:]=='fe')+'e'*(s[-1]in'sxz'or s[-2:]in('ch','sh')or s[-1]=='o'and s[-2]not in'aiueo')+'\bie'*(s[-1]=='y'and s[-2]not in'aiueo')+'s'

Rails runner, 18 bytes



$ echo knife | rails r filename.rb
  • \$\begingroup\$ Now that's an esoteric language. \$\endgroup\$ – Ven Mar 9 '17 at 9:24

Python, 296 bytes

z = input()
if z[-1]in['s','x','z','ch','sh']:print(z+'es')
elif z[-1]=='y'and z[-2]not in['a','e','i','o','u']:print(z[:-1]+'ies')
elif z[-2:]=='fe':print(z[:-2]+'ves')
elif z[-1]=='f':print(z[:-1]+'ves')
elif z[-1]=='o'and z[-2]not in['a','e','i','o','u']:print(z[:-1]+'oes')

C, 452 bytes

#define R return
#define I if
#define C unsigned char
#define F for
#define Q static
#define E else
C*p(C*b){Q C r[256],i,w,n;I(!b)R 0;F(i=0;i<250&&isalpha(r[i]=(b[i]|32));++i);I(i==250||i==0)R 0;w=r[i-1];n=i<2?0:r[i-2];I(strchr("sxz",w)||(w=='h'&&(n=='c'||n=='s'))||(w=='o'&&!strchr("aeiou",n)))r[i++]='e';E I(w=='y'&&!strchr("aeiou",n))r[i-1]='i',r[i++]='e';E I(w=='f')r[i-1]='v',r[i++]='e';E I(n=='f'&&w=='e')r[i-2]='v';r[i++]='s';r[i]=0;R r;} 

input as ascii only strings. Ungolf it and the main for test

     {Q C r[256],i,w,n;
      I(!b)R 0;
      I(i==250||i==0)R 0;
      E I(w=='y'&&!strchr("aeiou",n))
      E I(w=='f')r[i-1]='v',r[i++]='e';
      E I(n=='f'&&w=='e')
      r[i++]='s'; r[i]=0;
      R  r;
C *arst[]={"car","bus","potato","knife","penny","exception","wolf","eye","decoy","radio"};
#define P printf

//1100001 == 'a'  
//1000001 == 'A'
// 100000 ==  32
 {C  *b;
  int i;

      P("%s %s\n", arst[i], p(arst[i]));
  R 0; 


car cars
bus buses
potato potatoes
knife knives
penny pennies
exception exceptions
wolf wolves
eye eyes
decoy decoys
radio radios
  • \$\begingroup\$ It should be wolves not wolfves. \$\endgroup\$ – mbomb007 Apr 6 '17 at 18:07

Java 7, 408 bytes


boolean b="bcdfghjklmnpqrstvwxyzs".contains(String.valueOf(s.charAt(s.length()-2))); String x=s.substring(0,s.length()-1);if(s.endsWith("s")||s.endsWith("x")||s.endsWith("z")||s.endsWith("ch")||s.endsWith("sh"))return s+"es";if(s.endsWith("y")&&b)return x+"ies";if(s.endsWith("f")) return x+"ves";if(s.endsWith("fe"))return s.substring(0,s.length()-2)+"ves";if(s.endsWith("o")&&b)return s+"es";return s+="s";

Basically testing what the end of the String is and adding / replacing letters depending on what case it is. The boolean and String at the beginning are just for removing repetition in the test cases and making the code smaller.

Readable version:

public static String pluralize(String s){

// Consonant at the 2nd last position?
boolean b = "bcdfghjklmnpqrstvwxyzs".contains(String.valueOf(s.charAt(s.length()-2))); 

// Substring for cases where last letter needs to be replaced
String x = s.substring(0,s.length()-1);

if(s.endsWith("s") || s.endsWith("x") || s.endsWith("z") || s.endsWith("ch") || s.endsWith("sh"))
    return s + "es";
if(s.endsWith("y") && b)
    return x + "ies";
    return x + "ves";
    return s.substring(0,s.length()-2) + "ves";
if(s.endsWith("o") && b)
    return s + "es";

return s += "s";
  • 4
    \$\begingroup\$ You can't use a snippet. \$\endgroup\$ – Okx Mar 6 '17 at 10:10

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