381
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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)

      v
    creation
    
  • Find the first consonant after that

        v
    creation
    
  • Remove the rest of the string

    creat
    

That is your first sound group.

Next step:

Get the last consonant of the sound group

t

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

d

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:

didi

Concatenate this to the first sound group:

creatdidi

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!

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8
  • 7
    \$\begingroup\$ "x" should technically map onto "gz". "qu" should map onto "gw". \$\endgroup\$ Jun 5 '17 at 5:34
  • 78
    \$\begingroup\$ Answers over 140 characters should be disqualified \$\endgroup\$ Jun 7 '17 at 15:27
  • 16
    \$\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
  • 10
    \$\begingroup\$ @ThePlasmaRailgun It was a joke, since tweets have to be 140 characters or less. \$\endgroup\$ Dec 31 '17 at 22:14

44 Answers 44

1
2
2
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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:

creation->creatdidi
coverage->covfefe
example->exxaxa
programming->progkaka
code->codtete
president->preszizi
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2
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sed, 106 (105+1) bytes

This is sed with the -E flag, which apparently counts for one byte.

s/([aoeuiy][^aoeuiy])[^aoeuiy]*(.).*/\1\2/
h
s/.*(..)/\1\1/
y/bcdfgkpstvz/pgtvkgbzdfs/
x
s/.$//
G
s/\n//g

Try it online!

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2
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C#, 584 581 bytes

-3 bytes thanks to Destructible Lemon

This is my first submission on Code Golf and on Stack Exchange in general. I know that C# isn't a great golfing language and this probably isn't completely optimized but I wanted to give it a shot :p. Any tips are welcome!

Golfed Version:

namespace System{class B{static void Main(string[]args){var s="creation";var t="aeiou";int i=0,j=0,l=s.Length;char c=' ',f=' ';for(int x=0;x++<l;){if(t.IndexOf(s[x])>=0){i=x;break;}}for(int x=i;x++<l;){if(!(t.IndexOf(s[x])>=0)){j=x;c=s[x];for(int y=x;y++<l;){if (t.IndexOf(s[y])>=0){f=s[y];goto W;}}}}W:switch(c){case'b':c='p';break;case'c':c='g';break;case'd':c='t';break;case'f':c='v';break;case'g':c='k';break;case'k':c='j';break;case'p':c='b';break;case's':c='z';break;case't':c='d';break;case'v':c='f';break;case'z':c='s';break;}Console.Write(s.Substring(0,l-i-1)+c+f+c+f);}}}

Readable Version:

namespace System
{
    class B
    {
        static void Main(string[] args)
        {
            var s = "creation";
            var t = "aeiou";
            int i = 0, j = 0, l = s.Length;
            char c = ' ', f = ' ';
            for (int x = 0; x++ < l; )
            {
                if (t.IndexOf(s[x]) >= 0)
                {
                    i = x; break;
                }
            }
            for (int x = i; x++ < l;)
            {
                if (!(t.IndexOf(s[x]) >= 0))
                {
                    j = x; c = s[x];
                    for (int y = x; y++ < l;)
                    {
                        if (t.IndexOf(s[y]) >= 0)
                        {
                            f = s[y];
                            break;
                        }
                    }
                }
            }
            switch (c)
            {
                case 'b': c = 'p';
                    break;
                case 'c': c = 'g';
                    break;
                case 'd': c = 't';
                    break;
                case 'f': c = 'v';
                    break;
                case 'g': c = 'k';
                    break;
                case 'k': c = 'j';
                    break;
                case 'p': c = 'b';
                    break;
                case 's': c = 'z';
                    break;
                case 't': c = 'd';
                    break;
                case 'v': c = 'f';
                    break;
                case 'z': c = 's';
                    break;
            }
            Console.Write(s.Substring(0, l - i - 1) + c + f + c + f);
        }
    }
}
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2
  • 1
    \$\begingroup\$ I'm no expert, but I think you can add the incrementation to the comparator in the for loop, that is, x++ < l, or something (maybe l > x++ if the first doesn't work). not sure though \$\endgroup\$ Jul 22 '17 at 2:51
  • \$\begingroup\$ @DestructibleLemon Thanks for the tip! \$\endgroup\$ Jul 24 '17 at 14:18
2
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Vim, 107 keystrokes

Who needs Java, Python 3, Modern Pascal 2.0, C#, Python 2, R, Go, C, BlitzMax, Javascript, Crystal, Clojure, Lua, Matlab/Octave, Haskell and PHP when you have vim?

i ⎋o⏎bcdfghjklmnpqrstvwxz⏎pgtvkhjglmnbqrzdfwxs⎋1Ghqy/[aeiouy]⏎q/[^aeiouy]⏎mz@yyl`zpld$yhjpg*jyl`zpy2lPjdGX

is the Escape key and is the Return key

Explanation

i ⎋                         Insert a space before the first character
o⏎bcdfghjklmnpqrstvwxz⏎     Insert the character data
  pgtvkhjglmnbqrzdfwxs⎋1Gh
qy/[aeiouy]⏎q               Find the first vocal after the space
/[^aeiouy]⏎mz               Find the next consonant and add a marker
@yyl`zp                     Find the next vocal and put it after the consonant 
ld$                         Delete the rest of the world
yhjpg*                      Search for the consonant in the first row of the character data
jyl                         Copy the character in the same position in the second row
`zpy2lPjdGX                 Paste it after the last vowel and repeat the two last characters
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1
  • \$\begingroup\$ I may be wrong, but it doesn't look like [aeiou] and [^aeiou] account for the fact that, in this challenge, y is considered a vowel. \$\endgroup\$
    – FlipTack
    Jan 3 '18 at 16:32
2
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SmileBASIC, 185 179 168 bytes

This is an even more golfed version of snail_'s answer.

INPUT W$DEF V(C)RETURN.>INSTR(@AEIOUY,W$[C])END@L?W$[I];
I=I+1ON!V(I-1)&&V(I)GOTO@L
X$=W$[I]WHILE V(I)I=I+1WEND?X$;((X$+@PGTVKGBZDFS)[INSTR(@BCDFGKPSTVZ,X$)+1]+W$[I])*2

Ungolfed:

'input
INPUT WORD$
'function returns true if the character at POS is a consonant
DEF CONSONANT(POS)
 RETURN INSTR(@AEIOUY,WORD$[POS])<0
END
'scan through the word until there's a vowel followed by a consonant
REPEAT
 PRINT WORD$[I];
 I=I+1
UNTIL !CONSONANT(I-1) && CONSONANT(I)
'Store and print the final letter of the "normal" section
LASTNORMAL$=WORD$[I]
PRINT LASTNORMAL$;
'Find the next vowel
WHILE CONSONANT(I)
 I=I+1
WEND
'If last normal letter is one of "BCDFGKPSTVZ", replace with corresponding letter in "PGTVKGBZDFS"
'otherwise don't change it
CORRUPTED$=(LASTNORMAL$+@PGTVKGBZDFS)[INSTR(@BCDFGKPSTVZ,LASTNORMAL$)+1]
'print the "corrupted" letter and the vowel that was found earlier, twice.
PRINT (CORRUPTED$+WORD$[I])*2
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2
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Python 3, 170 149 bytes

def f(s):g=lambda x,y:[(i in"aeiouy")for i in s].index(x,y);j=g(0,g(1,0));c=s[j];z="ckbdfszvtpgg"+c;return s[:j+1]+2*((c+z)[-z.index(c)-2]+s[g(1,j)])

Try it online!

A more readable version:

def f(s):
    n = [ (i in "aeiou") for i in s ]
    i = n.index(1)
    j = n.index(0, i)
    k = n.index(1, j)
    c = s[j]
    v = s[k]
    z = "ckbdfszvtpgg" + c
    r = (c+z)[-z.index(c)-2]
    return s[:j+1] + r + v + r + v

i, j, and k are the indices of the first vowel, the first consonant after that c, and the first vowel after that v respectively.

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  • 2
    \$\begingroup\$ Welcome to the site and nice first answer! I've added in a link to an online interpreter so that others can test and verify your code. Note that you can save a few bytes by changing the indentation to a single space on each line. For more tips, check out our Tips for golfing in Python page! \$\endgroup\$ Mar 10 '21 at 20:53
  • 2
    \$\begingroup\$ Welcome to the site! More than just a single space, you can actually get rid of all of the indentation by using ;. You also have an extra space after the first in. Try it online! \$\endgroup\$
    – Wheat Wizard
    Mar 10 '21 at 20:55
  • \$\begingroup\$ Thank you guys for the optimizations! \$\endgroup\$
    – Roy
    Mar 11 '21 at 21:22
2
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Python 3, 190 bytes

f=lambda s,g='aeiouy':min([s.index(l)for l in g if l in s])
C='bcdfghjklmnpqrstvwxyz'
q=input()
v=f(q)
c=f(q[v:],C)+v+1
print(q[:c]+('pgtvkhjglmnbqrzdfwxs'[C.index(q[c-1])]+q[f(q[c:])+c])*2)

Try it online!

I've had this coded since the week of the challenge, but was discouraged to post in its then-current state (which also happens to be the current state) because of all the 80-byte regex answers.

But hey, what the hell. Here's another non-esoteric, non-regex answer, six months late to the party.

-30 bytes from typing out the consonants in a string rather than using a list comprehension

-4 bytes from using the result of a calculation inline rather than assigning it to a variable (whoops)

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2
  • 1
    \$\begingroup\$ Your TIO gives an error. \$\endgroup\$
    – FlipTack
    Jan 4 '18 at 7:49
  • \$\begingroup\$ @FlipTack turns out I'm just an idiot - "test" is not a valid test string. Updated the link and shaved 4 bytes off of the source \$\endgroup\$
    – osuka_
    Apr 21 '21 at 5:42
2
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PowerShell, 324 bytes

$v="[aeiouy]";$f=[regex]::match($w,$v).index+1;$s=[regex]::match($w.substring($f),"[^aeiouy]").index;$t=[regex]::match($w.substring($f),$v);
$c=@{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'};
$r=$c["$($w[($f+$s)])"];$w.substring(0,$f+$s+1)+"$r$t$r$t"

This was as short as I could get it without frying my brain. Takes input as $w='coverage'.

Try it online!

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Perl, 90 bytes (89 + 1)

I see that I've already been beaten by another Perl answer, but I'll post this anyway. Run with -p.

s/[aeiouy](.)\K[^aeiouy]*(.).*$//;$x=$1=~y/bcdfgkpstvz/pgtvkgbzdfs/r;$_.="$x$2"x2
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1
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Go, 298 bytes

func c(l string)string{v:="aeiouy";c:="bcdfghjklmnpqrstvwxz";r:="pgtvkhjglmnbqrzdfwxs";b:=l[0:strings.IndexAny(l,v)+1];t:=strings.SplitAfterN(l,b,2)[1];i:=strings.IndexAny(t,c);b=b+t[0:i+1];x:=string(r[strings.IndexAny(c,string(t[i]))]);z:=string(t[i:][strings.IndexAny(t[i:],v)]);return b+x+z+x+z}
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1
  • \$\begingroup\$ Welcome to PPCG! \$\endgroup\$ Jun 19 '17 at 14:46
1
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Modern Pascal 2.0, 415, 396 bytes

function c(S:String):String;
var L,M,I:Longint; const v='aeiouy'; o='_pgt_vkh_jglmn_bqrzd_fwx_s';
begin S:=Lowercase(S); Result:=S;
For l:=1 to length(S) do if pos(S[l],v)>0 then Break;
For M:=l to length(S) do if pos(S[M],v)=0 then begin
Result:=Copy(S,1,M);I:=Ord(S[M])-96;Delete(S,1,M);
For L:=1 to Length(S) do If Pos(S[l],v)>0 then begin
Result+=o[I]+S[L]+o[I]+S[L];Break;End;Break;End;End;

// Author of Modern Pascal

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3
  • 1
    \$\begingroup\$ this is not 415 bytes. please trim spaces and actually make it 415 bytes to have that score. also the function name can be golfed \$\endgroup\$ Jun 12 '17 at 23:52
  • \$\begingroup\$ made smaller by unformatting. Could go even smaller, as modern pascal does not enforce the grammer of then a wasted heartbeat in parsing. \$\endgroup\$
    – Ozz Nixon
    Jun 13 '17 at 17:46
  • 3
    \$\begingroup\$ if you can go even smaller then you should \$\endgroup\$ Jun 13 '17 at 22:17
1
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BQN, 79 bytesSBCS

{s∾4⥊("s_xwf_dzrqb_nmlgj_hkv_tgp"⊑˜'z'-⊑⌽s←⊑w⊔𝕩)∾⊑𝕩/˜v∧w←0∾»∨`1‿0⍷v←𝕩∊"aeiouy"}

Run online!

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1
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Lexurgy, 288 bytes

Docs here.

Lexurgy is an online tool meant for conlangers (people who make constructed languages) to apply sound changes to their conlangs via a series of programmable rules. As such, this tool involves many string operations.

Place input in the left box, code in the center, and click "Apply" to view the output in the right box.

(Note: Lexurgy is a programming language according to our definition of a programming language, since Lexurgy's provided examples shows a functional adding machine implemented in Lexurgy. Theoretically, one could take a unary primality test and implement it in Lexurgy as well.)

Class v {a,e,i,o,u}
Class u {p,t,k,f,s,c}
Class w {b,d,g,v,z,g}
Class c {@u,@w,h,j,l,m,n,q,r,w,x}
a:
@c *=>@c ;/$ @c* @v* _//{$ _,$ @c _}
b:
@c=>* /; _
c:
{@v,@c}=>* /; @v {@v, @c}* _
d:
* @w$1=>$1 @u/_ ;
* @u$1=>$1 @w/_ ;
e:
@c$1 ;=>; $1
f:
(@c @v)$1=>$1 $1/; _
g:
;=>*

Ungolfed:

Class vowel {a,e,i,o,u}
Class unvoiced {p,t,k,f,s,c}
Class voiced {b,d,g,v,z,g}
Class consonant {@unvoiced,@voiced,h,j,l,m,n,q,r,w,x}

# find first consonant after first vowel
part-1:
 @consonant * => @consonant ; / $ @consonant* @vowel* _ // {$ _, $ @consonant _}

romanizer-a:
 unchanged

# delete the consonant after the seperator
part-2:
 @consonant => * / ; _

romanizer-b:
 unchanged

# remove everything except the second vowel
part-3:
 {@vowel, @consonant} => * / ; @vowel {@vowel, @consonant}* _

romanizer-c:
 unchanged

# voicings
part-4:
 * @voiced$1 => $1 @unvoiced / _ ;
 * @unvoiced$1 => $1 @voiced / _ ;

romanizer-d:
 unchanged

# swap mapped consonant and seperator
part-5:
 @consonant$1 ; => ; $1

romanizer-e:
 unchanged

# duplicate the `fe`
part-6:
 (@consonant @vowel)$1 => $1 $1 / ; _

romanizer-f:
 unchanged

# remove the seperator
part-7:
 ; => *
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0
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Excel, 207 bytes

=LET(s,A1,x,SEQUENCE(LEN(s)),c,ISERROR(FIND(MID(s,x,1),"aeiouy")),k,MIN(IF(c*(x>MIN(IF(c,"",x))),x,"")),q,MID(s,k,1),m,"bcdfkszgvtgp"&q,n,MID(q&m,14-FIND(q,m),1)&MID(s,MIN(IF(c+(x<k),"",x)),1),LEFT(s,k)&n&n)
\$\endgroup\$
1
2

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