16
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The challenge is simple:

generate a word.

Specifications:

  • Word must be pronounceable.
    • This is defined as "alternating between a consonant and a vowel."
    • A consonant is one of the following letters: bcdfghjklmnpqrstvwxz
    • A vowel is one of the following letters: aeiouy
  • Word must be randomly generated.
  • Words must be able to contain every consonant and vowel. (You can't just use bcdf for consonants and aei for vowels.)
  • Word must contain 10 letters.
  • Shortest code (in character count) wins.
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  • 1
    \$\begingroup\$ codegolf.stackexchange.com/questions/11215/alien-name-generator \$\endgroup\$ – John Dvorak Jun 18 '13 at 4:43
  • 7
    \$\begingroup\$ With this xkcd strip in mind, the program echo buxitiwymu technically conforms to the specification. I assure you, I generated the word randomly:P \$\endgroup\$ – AardvarkSoup Jun 23 '13 at 22:02
  • 1
    \$\begingroup\$ @AardvarkSoup "Words must be able to contain every consonant and vowel" \$\endgroup\$ – Doorknob Jun 23 '13 at 22:52
  • 1
    \$\begingroup\$ @Kartik depends on the context, in 'yes' it's a consonant, in 'why' it's a vowel, but this would make it impossible to define a pronounceable word as alternating between vowels and consonants, eg. yyyyyyyy would be a valid word. \$\endgroup\$ – CJStuart Feb 17 '15 at 22:30
  • 1
    \$\begingroup\$ I actually made a generator on Scratch a while back. It had specific rules for when you can treat y as a vowel, where you can use q and x, and when you can use two-letter combinations like ng or ea \$\endgroup\$ – Esolanging Fruit Nov 6 '16 at 21:42

27 Answers 27

7
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GolfScript, 32 characters

['aeiouy'.123,97>^]5*{.,rand=}%+

Run it online.

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8
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Ruby: 56 characters

([v=%w(a e i o u y),[*?a..?z]-v]*5).map{|a|$><<a.sample}

Example outputs:

  • itopytojog
  • umapujojim
  • ipagodusas
  • yfoqinifyw
  • ebylipodiz
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  • 1
    \$\begingroup\$ 'aeiouy'.chars would be one char shorter. \$\endgroup\$ – Howard Jun 19 '13 at 6:01
  • \$\begingroup\$ @Howard then the subtraction operator raises TypeError: can't convert Enumerator into Array \$\endgroup\$ – John Dvorak Jun 19 '13 at 6:30
  • \$\begingroup\$ @JanDvorak Sorry, forgot to mention that you need Ruby 2.0 for this trick. \$\endgroup\$ – Howard Jun 19 '13 at 6:38
7
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Python, 81

from random import*
print''.join(map(choice,["bcdfghjklmnpqrstvwxz","aeiouy"]*5))

Good luck pronouncing them.

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  • \$\begingroup\$ They're actually quite easy to pronounce, for example the first word I got was "viketuziwo" :P \$\endgroup\$ – Doorknob Jun 18 '13 at 3:16
  • \$\begingroup\$ @Doorknob maybe it's just my luck. I keep getting 'words' like "qijepyjyga". My computer's attempts to pronounce them make up for it though :) \$\endgroup\$ – grc Jun 18 '13 at 5:04
  • 3
    \$\begingroup\$ I just had a lot of fun doing python grc.py | say on my machine. Thanks for the idea. \$\endgroup\$ – Kaya Jun 21 '13 at 3:07
6
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APL (34)

,⍉↑(V[5?6])((⎕A~V←'AEIOUY')[5?20])
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6
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JavaScript, 74

for(a=b=[];a++-5;)b+="bcdfghjklmnpqrstvwxz"[c=new Date*a%20]+"aeiouy"[c%6]

Does not generate all combinations, but I think that all consonants and vowel appear.

JavaScript, 79

for(a=b=[];a--+5;)b+="bcdfghjklmnpqrstvwxz"[c=Math.random()*20^0]+"aeiouy"[c%6]

More "random" version.

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  • \$\begingroup\$ What's the ^0 for? \$\endgroup\$ – Alpha Jun 25 '13 at 3:01
  • 1
    \$\begingroup\$ @Alpha Math.random gives a float, we need an integer. ^0 truncates the number \$\endgroup\$ – copy Jun 25 '13 at 11:52
  • \$\begingroup\$ Clever trick. I haven't seen the ^ operator in JavaScript before and less heard about using it to truncate a float. Thanks! \$\endgroup\$ – Alpha Jun 25 '13 at 17:06
  • \$\begingroup\$ @copy i like the use of ^ \$\endgroup\$ – Math chiller Oct 2 '13 at 3:16
6
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COBOL, 255

So I'm learning COBOL at the moment. Used this question as some practice. Tried to golf it.

It's 255 without the leading whitespace, and 286 bytes with.

For what it's worth, this runs in Microfocus COBOL for VS2012, and I have no idea if it will run anywhere else.

       1 l pic 99 1 s pic x(26) value'aeiouybcdfghjklmnpqrstvwxz' 1 r
       pic 99 1 v pic 9 1 q pic 99. accept q perform test after varying
       l from 1 by 1 until l>9 compute v=function mod(l,2) compute r=1+
       function random(q*l)*5+v*15+v*5 display s(r:1)end-perform exit
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4
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Ruby: 70 66 characters

10.times{|i|$><<["aeiouy"*4,"bcdfghjklmnpqrstvwxz"][i%2][rand 20]}

Sample run:

bash-4.1$ ruby -e '10.times{|i|$><<["aeiouy"*4,"bcdfghjklmnpqrstvwxz"][i%2][rand 20]}'
izogoreroz

bash-4.1$ ruby -e '10.times{|i|$><<["aeiouy"*4,"bcdfghjklmnpqrstvwxz"][i%2][rand 20]}'
onewijolen

bash-4.1$ ruby -e '10.times{|i|$><<["aeiouy"*4,"bcdfghjklmnpqrstvwxz"][i%2][rand 20]}'
amilyfilil
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  • \$\begingroup\$ You can use 10.times for one char less. \$\endgroup\$ – Howard Jun 19 '13 at 4:06
  • 1
    \$\begingroup\$ Also the question doesn't require that each letter must have same probability. *10->*4, skip *3, rand 60->rand 20 and you have saved 3 chars. \$\endgroup\$ – Howard Jun 19 '13 at 4:10
  • \$\begingroup\$ Good catch on the rule, @Howard. Thank you. \$\endgroup\$ – manatwork Jun 19 '13 at 7:18
4
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R: 105 characters

a=c(1,5,9,15,21,25)
l=letters
s=sample
cat(apply(cbind(s(l[-a],5),s(l[a],5)),1,paste,collapse=""),sep="")
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4
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J (51)

,|:>(<"1[5?6 20){&.>'aeiouy';'bcdfghjklmnpqrstvwxz'
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4
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Processing, 100 99 93 87

int i=10;while(i-->0)"aeiouybcdfghjklmnpqrstvwxz".charAt((int)random(i%2*6,6+i%2*i*2));

Upon closer inspection of the question, I see it doesn't require any output. I've adjusted this accordingly.

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3
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Objective-C, 129

main(){int i=10;while(i-->0)printf("%c",[(i%2==0)?@"aeiouy":@"bcdfghjklmnpqrstvwxz"characterAtIndex:arc4random()%(6+(i%2)*14)]);}

With the help of Daniero

(I love to use the tends to operator (-->)

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3
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Java AKA the most verbose language ever created, 176 with help of Doorknob, Daniero and Peter Taylor (thanks guys!)

class w{public static void main(String[]a){int i=11;while(--i>0)System.out.print((i%2==0?"aeiouy":"bcdfghjklmnpqrstvwxz").charAt(new java.util.Random().nextInt(6+(i%2)*14)));}}

Ungolfed:

    class w {

        public static void main(String[] a) {
            int i = 11;
            while (--i > 0) {
                System.out.print((i % 2 == 0 ? "aeiouy" : "bcdfghjklmnpqrstvwxz").charAt(new java.util.Random().nextInt(6 + (i % 2) * 14)));
            }
     }

}

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  • 1
    \$\begingroup\$ Suggested improvements: Change String a[] to String[]a (-1 characters), change w W = new w(); to w W=new w(); (-2 characters) \$\endgroup\$ – Doorknob Jun 18 '13 at 22:14
  • \$\begingroup\$ Suggestions: Let the word always start on a consonant (or wovel); no need to randomize this when the question doesn't mention it! So, skip the boolean f and use i%2 instead. Also, the for loop can be shortened, and you can put both the strings inside the conditional operator (also,no need for parens here), and use the charAt on the outside. Here's the whole thing, 195 CHARS, 38 SAVED: import java.util.*;class w{public static void main(String[]a){Random r=new Random();for(int i=0;++i<11;)System.out.print((i%2>0?"bcdfghjklmnpqrstvwxz":"aeiouy").charAt(r.nextInt(6+(i%2)*14)));}} \$\endgroup\$ – daniero Jun 19 '13 at 1:40
  • 2
    \$\begingroup\$ “Java AKA the most verbose language ever created” – Ever tried Shakespeare? ;-) \$\endgroup\$ – manatwork Jun 19 '13 at 7:13
  • 1
    \$\begingroup\$ @manatwork, don't forget to remove the import once you've got it down to being used only in one place. \$\endgroup\$ – Peter Taylor Jun 19 '13 at 11:52
  • 1
    \$\begingroup\$ @manatwork maybe we like lisp ?? ok, sorry, gonna take those parenthesis out when I get home from work \$\endgroup\$ – jsedano Jun 20 '13 at 15:18
3
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Javascript, 85

for(r=Math.random,s="",i=5;i--;)s+="bcdfghjklmnpqrstvwxz"[20*r()|0]+"aeiouy"[6*r()|0]

If run from the console, output is shown. Explicit display would add alert(s) at 8 chars, still shorter than the other JS solutions.

Thanks C5H8NNaO4 and Howard!

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  • \$\begingroup\$ Nice one, save a character by removing the last ';' \$\endgroup\$ – C5H8NNaO4 Jun 18 '13 at 11:20
  • 1
    \$\begingroup\$ Instead of ~~(###) you can write ###|0 which saves 4 chars. \$\endgroup\$ – Howard Jun 19 '13 at 12:44
3
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Javascript, 135 122 96 86 characters

s='';c=5;r=Math.random;while(c--)s+='bcdfghjklmnpqrstvwxz'[r()*20|0]+'aeiouy'[r()*6|0]
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3
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PHP, 100 Characters

<?while($i<6){echo substr('bcdfghjklmnpqrstvwxz',rand(0,20),1).substr('aeiouy',rand(0,5),1);$i++;}?>
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  • \$\begingroup\$ acually I think that's 100 chars long \$\endgroup\$ – Cristian Lupascu Jul 24 '13 at 11:27
3
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Pyth, 26 characters

J-G"aeiouy"VT=k+kOJ=J-GJ;k

You can try it in the online compiler here.

Someone posted a very similar challenge but it was closed after I had made a solution. I didn't realize it, but this question actually predates the creation of Pyth. Anyway, here is the breakdown:

J                             Set string J equal to:
  G                            the entire alphabet (lowercase, in order)
 - "aeiouy"                    minus the vowels
           VT                 For n in range(1, 10):
             =k                   Set string k to:
                k                  k (defaults to empty string)
               + OJ                plus a random character from J
                   =J             Set J to:
                      G            the entire alphabet
                     - J           minus J
                        ;     End of loop
                         k    print k

Every time the loop is run, J switches from being a list of consonants to a list of vowels. That way we can just pick a random letter from J each time.
There may be a way to initialize J in the loop or remove the explicit assignments from the loop, but I have not had success with either yet.

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  • 1
    \$\begingroup\$ Bumping old questions is generally considered fine. However, when you use a language newer than the question (I created Pyth in 2014) you should note this in your answer. \$\endgroup\$ – isaacg Jul 7 '15 at 3:42
  • \$\begingroup\$ Thanks for clearing that up for me. I didn't realize that Pyth was created after this question and I've added that to the answer. \$\endgroup\$ – Mike Bufardeci Jul 7 '15 at 4:11
3
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APL 30 26

,('EIOUY'∪⎕A)[6(|,+)⍪5?20]

Explanation is very similar to the past version below, just reordered a bit to golf the solution.

Note: ⎕IO is set to 0


('EIOUY'∪⎕A)[,6(|,(⍪+))5?20]

Explanation:

'EIOUY'∪⎕A    puts vowels in front of all letters.
5?20            for the indexes we start choosing 5 random numbers between 0 and 19
6(|,(⍪+))        then we sum 6 and the random numbers, convert to 5x1 matrix (⍪), add a column before this one containing 6 modulo the random numbers. 
                [[[ this one can be rewritten as: (6|n) , ⍪(6+n)  for easier understanding]]]
,6(|,(⍪+))5?20  the leading comma just converts the matrix to a vector, mixing the vowel and consonants indexes.

Tryapl.org

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  • 1
    \$\begingroup\$ ('AEIOUY'∪⎕A) ≡ (∪'AEIOUY',⎕A) but it's one byte shorter. \$\endgroup\$ – lstefano Jun 30 '16 at 12:42
  • 1
    \$\begingroup\$ Deferring 'A' to ⎕A saves another byte: ,('EIOUY'∪⎕A)[6(|,+)⍪5?20] \$\endgroup\$ – Adám Jul 3 '16 at 4:03
  • \$\begingroup\$ Nice! down to 26. Thanks \$\endgroup\$ – Moris Zucca Jul 4 '16 at 7:54
2
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PHP 79 bytes

<?for($c=bcdfghjklmnpqrstvwxz,$v=aeiouy;5>$i++;)echo$c[rand()%20],$v[rand()%6];

Fairly concise.

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2
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C: 101

main(){int x=5;while(x-->0){putchar("bcdfghjklmnpqrstvwxyz"[rand()%21]);putchar("aeiou"[rand()%5]);}}
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2
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Javascript, 104 87

a="";for(e=10;e--;)a+=(b=e&1?"bcdfghjklmnpqrstvwxz":"aeiouy")[0|Math.random()*b.length]

golfed a whole lot of simple unnecessary stuff, still not nearly as nice as copys' one

Oh, and that one just opped up during golfing: "dydudelidu"

Now I tried one using the 2 characters at once approach. Turns out it's almost the same as copys' second one, so I can't count it, also at 79. a="";for(e=5;e--;)a+="bcdfghjklmnpqrstvwxz"[m=0|20*Math.random()]+"aeiouy"[m%6]

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2
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Unix tools: 73 bytes

And not guaranteed running time :)

</dev/urandom grep -ao '[aeiouy][bcdfghjklmnpqrstvwxz]'|head -5|paste -sd ''

Only problem is that the generated string will start with a "vowel" every time.

(edit: changed ' ' to '' in the args of paste) (another edit: removed -P from grep options, thanks to manatwork)

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  • \$\begingroup\$ Something could be fine tuned around the grep parameters. I got “of e� ap ag ak”. \$\endgroup\$ – manatwork Feb 4 '15 at 13:56
  • \$\begingroup\$ Hmm... strange. I got nothing like that. I thought -a would be enough. \$\endgroup\$ – pgy Feb 4 '15 at 13:58
  • 1
    \$\begingroup\$ My test indicates that -P is the one. Seems the man warns about its highly experimental status with a reason. (grep 2.16) But anyway, it works fine without -P. \$\endgroup\$ – manatwork Feb 4 '15 at 14:01
  • \$\begingroup\$ You are right thank you, I didn't consider that. I don't even know why I used -P in the first place. I'll edit my answer. \$\endgroup\$ – pgy Feb 4 '15 at 14:03
  • 1
    \$\begingroup\$ By the way, tr -d \\n is shorter for joining the lines. \$\endgroup\$ – manatwork Feb 4 '15 at 14:04
2
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Brachylog, 9 bytes

Ḍ;Ẉṣᵐzh₅c

Try it online!

Gives output as a list of letters through the output variable.

   ṣ         Randomly shuffle
 ;  ᵐ        both
Ḍ            the consonants without y
  Ẉ          and the vowels with y,
     z       zip them into pairs,
      h₅     take the first five pairs,
             and output
        c    their concatenation.
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1
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F#, 166 characters

open System;String.Join("",(Random(),"aeiouybcdfghjklmnpqrstvwxz")|>(fun(r,s)->[0..5]|>List.collect(fun t->[s.Substring(6+r.Next()%20,1);s.Substring(r.Next()%6,1)])))
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1
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K, 40 bytes

,/+5?/:("bcdfghjklmnpqrstvwxz";"aeiouy")

5?"abc" will generate 5 random letters from the given string.

5?/: will generate 5 random letters from each of the strings on the right, producing two lists.

+ transposes those two lists, giving us a list of tuples with one random character from the first list and then one from the second list.

,/ is "raze"- fuse together all those tuples in sequence.

K5 can do this in 33 bytes by building the alphabet more cleverly and then using "except" (^) to remove the vowels, but K5 is much too new to be legal in this question:

,/+5?/:((`c$97+!26)^v;v:"aeiouy")
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1
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R, 83 bytes

cat(outer((l<-letters)[a<-c(1,5,9,15,21,25)],l[-a],paste0)[sample(1:120,5)],sep="")

Generate all possible vowel-consonant sequences in a matrix, then randomly sample 5 of them, yielding a 10-letter word.

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1
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05AB1E, 10 bytes

5FžPΩ?žOΩ?

Try it online or output \$n\$ amount of random words.

Explanation:

Pretty straight-forward:

5F      # Loop 5 times:
  žP    #  Push string "bcdfghjklmnpqrstvwxz"
    Ω   #  Pop and push a random character from this string
     ?  #  Output it without newline
  žOΩ?  #  Do the same with string "aeiouy"
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0
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Jelly, 13 bytes

Øy,ØYẊ€Zs5ḢŒl

Explanation:

Øy,ØYẊ€Zs5ḢŒl
Øy,ØY         Makes a list of two lists of characters: [[list-of-vowels-with-y], [list-of-consonants-without-y]]
     Ẋ€       Shuffle €ach.
       Z      Zip those two lists together. [[random-vowel, random-consonant],...]
        s5    Split them into chunks of five (because each pair contains two letters, this splits them into chunks of 10 letters)
          Ḣ   Take the list of 5 pairs.
           Œl Make each of the characters in it lowercase
              Implicit concatenation and print.

Try it online!

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