55
\$\begingroup\$

According to some controversial story, the odrer of ltteres in a wrod deos not mttaer much for raednig, as lnog as the frist and lsat lteter macth with the orignial wrod.

So, for fun, what would be the shortest function to randomize letter order in a word while keeping the first and the last letter in place?

Here's my stab at it with JavaScript. All whitespace removed it's at 124 130 characters.

function r(w) {
  var l=w.length-1;
  return l<3?w:w[0]+w.slice(1,l).split("").sort(function(){return Math.random()-.5}).join("")+w[l];
}

Shorter JavaScript always welcome.


  • Edit: length check added. Function should not fail for short words.
\$\endgroup\$
15
  • 3
    \$\begingroup\$ Haskell, 4 characters: r=id. \$\endgroup\$ Aug 3 '11 at 20:01
  • 2
    \$\begingroup\$ Yes. It returns the exact same thing as the input. On another note, what do we do about punctuation? Do we only operate on words consisting only of letters? \$\endgroup\$ Aug 3 '11 at 20:28
  • 1
    \$\begingroup\$ @trinithis not sure what you talking about, but id is the identity function. I would still like to see Haskell solution to this problem in less than 100 characters. \$\endgroup\$
    – Arlen
    Aug 4 '11 at 7:37
  • 3
    \$\begingroup\$ Should the specification be updated to require a uniform distribution of outcomes? This would disallow the 4 character Haskell solution. It would also disallow your example Javascript solution (shuffling by doing a sort like that is not uniform). \$\endgroup\$ Aug 4 '11 at 17:25
  • 2
    \$\begingroup\$ +1 for the first sentence: it actually took me a few seconds to realize it was spelled wrong XP \$\endgroup\$ Aug 4 '11 at 18:33

71 Answers 71

21
\$\begingroup\$

Haskell, 110 120 107 characters

import Random
s l=randomRIO(1,length l-2)>>=g.($l).splitAt
g(a:b,c:d)=fmap(a:).s$c:b++d
g(a,b)=return$a++b

An example of a program using this function:

main = getLine >>= s >>= putStrLn
\$\endgroup\$
3
  • \$\begingroup\$ fmap((a:t!!i:).tail) \$\endgroup\$
    – FUZxxl
    Aug 5 '11 at 22:00
  • \$\begingroup\$ @FUZxxl, no, sorry, I made the same mistake initially. Operator section does not work here because : is right-associative. I'd have to do (a:).(t!!i:). \$\endgroup\$
    – Rotsor
    Aug 5 '11 at 22:08
  • \$\begingroup\$ ~3.5 more years of experience, and another 13 characters saved! \$\endgroup\$
    – Rotsor
    Feb 14 '15 at 21:56
19
\$\begingroup\$

J, 26 24 23 characters

r=:{.,({~?~@#)&}.&}:,{:
\$\endgroup\$
2
  • \$\begingroup\$ According to common code golf rules, you don't have to bind the phrase to a name. \$\endgroup\$
    – FUZxxl
    Feb 14 '15 at 22:51
  • \$\begingroup\$ #?# is one char shorter than?~@# \$\endgroup\$
    – randomra
    Feb 15 '15 at 22:11
15
\$\begingroup\$

Ruby, 44 characters

r=->w{w[h=1..-2]=[*w[h].chars].shuffle*"";w}

Works also for short words, i.e. words with one, two or three characters are returned unaltered.

Edit: Using the array-splat idea of Ventero saves another char.

\$\endgroup\$
2
  • \$\begingroup\$ That's actually 44 characters. I eventually came up with the exact same answer by fine tuning my own - now I feel like a copy-cat after reading yours. \$\endgroup\$ Aug 4 '11 at 8:24
  • \$\begingroup\$ @user2316 Of course you are right. Thank you. \$\endgroup\$
    – Howard
    Aug 4 '11 at 9:50
11
\$\begingroup\$

Ruby 1.9, 46 characters

r=->w{w[0]+[*w[1..-2].chars].shuffle*""+w[-1]}
\$\endgroup\$
3
  • \$\begingroup\$ +1 This use of array-splat saved me also one char. Great idea. \$\endgroup\$
    – Howard
    Aug 3 '11 at 16:46
  • \$\begingroup\$ I don't know ruby - it fails on my ruby1.8, so I guess I need a never version? Does it work with input like 'I'? \$\endgroup\$ Aug 4 '11 at 4:00
  • \$\begingroup\$ @user: It says "Ruby 1.9" right there. ;) -- Ventero - One of the sensible requirements I forgot to mention is that it should not fail for word lengths 0 and 1. Sorry. \$\endgroup\$
    – Tomalak
    Aug 4 '11 at 6:34
11
\$\begingroup\$

C++, 79 characters (with range check)

string f(string s){if(s.size()>3)random_shuffle(&s[1],&s.end()[-1]);return s;}

C++, 81 65 characters (without range check)

string f(string s){random_shuffle(&s[1],&s.end()[-1]);return s;}

Using pass by reference instead of returning the result shaves off another 10 characters from either solution.

Full program, reading a string of words and shuffling converting them:

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <ctime>
#include <string>

using namespace std;    
string f(string s){if(s.size()>3)random_shuffle(&s[1],&s.end()[-1]);return s;}

int main() {
    std::srand(std::time(0));
    std::string s;
    while(std::cin >> s)
        std::cout << f(s) << " ";
    std::cout << std::endl;
}

Morale: don’t build what’s already there. Oh, and overflow checks are for wusses.

\$\endgroup\$
9
  • \$\begingroup\$ Nice, std::random_shuffle, that's a new one to me. btw I think you forgot #include<string> in your full code. \$\endgroup\$ Aug 3 '11 at 15:48
  • 1
    \$\begingroup\$ Something like that is what I had in mind originally. Unfortunately there is no built-in for shuffling a string in-place in JS. \$\endgroup\$
    – Tomalak
    Aug 3 '11 at 15:59
  • \$\begingroup\$ Fails for very short strings. \$\endgroup\$ Aug 4 '11 at 4:02
  • \$\begingroup\$ That's true, you're missing a length check (I did, as well). BTW @Arlen's answer is also worth a look. \$\endgroup\$
    – Tomalak
    Aug 4 '11 at 6:43
  • 1
    \$\begingroup\$ @userunknown That’s what I meant by “overflow checks are for wusses”. But to be fair, so do almost all other solutions. \$\endgroup\$ Aug 4 '11 at 7:02
7
\$\begingroup\$

C (K&R) - 88 86 87 chars

r(char*s){int m,j,l=strlen(s)-2,i=l;while(--i>0){j=rand()%l+1;m=s[j];s[j]=s[1];s[1]=m;}}

There's no build-in swap or shuffle function in C, so I had to do it manually :(

Sample Program with Ungolfed r():

#include <stdio.h>
#include <string.h>
#include <time.h>
#include <stdlib.h>

// -----------------------------------------------------------------------
r( char *s )
{
    int m, j, l=strlen(s)-2, i=l;

    while (--i>0)
    {
        j = rand() % l + 1;

        m = s[j];
        s[j] = s[1];
        s[1] = m;
    }

}
// -----------------------------------------------------------------------
int main()
{
    char s[] = "anticipated";

    srand( time(0) );
    r( s );
    puts( s );

    return 0;
}

EDIT: fixed the bug when s consists of less than 3 chars (thanks to user-uknown for noticing it! )

\$\endgroup\$
4
  • 1
    \$\begingroup\$ In glibc, there is (was?) a non-standard library function strfry. \$\endgroup\$ Aug 3 '11 at 23:12
  • \$\begingroup\$ funny experience, if I feed it with char s[] = "na"; // not anticipated \$\endgroup\$ Aug 4 '11 at 3:30
  • \$\begingroup\$ @user uknown: I just edited the code and fixed it, thanks for noticing the bug! (I just added >0 in the condition of the while loop, "costing me" two more, but needed, chars :) ) \$\endgroup\$
    – Harry K.
    Aug 4 '11 at 7:53
  • 1
    \$\begingroup\$ @Mechanical snail: I think strfy is still in glibc. \$\endgroup\$
    – Harry K.
    Aug 4 '11 at 7:56
7
\$\begingroup\$

Python, 87 79 75 93 92 chars

from random import*
f=lambda w:w if 4>len(w)else w[0]+''.join(sample(w[1:-1],len(w)-2))+w[-1]

EDIT: Originally thought it was supposed to split string words (which it did at 128 chars; now at 87 chars does requirement). Argh, my bad at reading comprehension.

EDIT 2: Change from def to lambda function from def to save 6 chars. Assuming sample is already imported to the namespace (from random import sample) could bring this down to ~60).

EDIT 3: "len(w[1:-1])" (12 chars) to "len(w)-2" (8 chars) per gnibbler's nice suggestion.

EDIT 4: JBernando saved one char (had considered from random import * and saw it was equivalent -- not realizing the space in import * is unnecessary).; user unknown added 19 chars w if len(w)<4 else to handle 0 and 1 char strings correctly.

EDIT 5: Saved another char per boothby's code golf trick. if len(w)<4 else to if 4>len(w)else.

\$\endgroup\$
19
  • \$\begingroup\$ However, the question only defined the input as a word, not a string of words. :) \$\endgroup\$ Aug 3 '11 at 17:10
  • 1
    \$\begingroup\$ @sidran32: Thanks, my bad. I had just noticed (upon rereading) and then saw your comment; deleted -- edited -- and undeleted. \$\endgroup\$
    – dr jimbob
    Aug 3 '11 at 17:19
  • \$\begingroup\$ Idea - you can trim 3 chars by doing this.... def f(w):j=w[1:-1]; return w[0]+''.join(r.sample(j,len(j)))+w[-1] \$\endgroup\$
    – arrdem
    Aug 3 '11 at 20:12
  • \$\begingroup\$ @rmckenzie: Good idea. However, right before I saw your comment right after I trimmed it to lambda function (saving 6 chars), so I can no longer do your method of defining intermediate vars. \$\endgroup\$
    – dr jimbob
    Aug 3 '11 at 20:20
  • 3
    \$\begingroup\$ len(w)-2 instead of len(w[1:-1])? \$\endgroup\$
    – gnibbler
    Aug 3 '11 at 23:29
6
\$\begingroup\$

C++, 111 97 chars

std::string f(std::string s){for(int i=s.size()-1;i>1;std::swap(s[rand()%i+1],s[--i]));return s;}

Here is a full program for those who wish to test it:

#include<string>
#include<iostream>

std::string f(std::string s){for(int i=s.size()-1;i>1;std::swap(s[rand()%i+1],s[--i]));return s;}

int main(){
    for(int i = 0; i<100; ++i)
    std::cout<<f("letters")<<std::endl;
}

Edit

Realised there is no need to random both swap indexes, saved a variable and a few more characters.

\$\endgroup\$
1
  • \$\begingroup\$ Excellent. Most solutions fail on very small input. Yours not. \$\endgroup\$ Aug 4 '11 at 4:06
6
\$\begingroup\$

php (68 characters)

$r=preg_replace('/^(\w)(\w+)(\w)$/e','$1.str_shuffle($2).$3',trim($w));

shorter (60 characters)

$r=preg_replace('/(.)(.+)(.)/e','$1.str_shuffle($2).$3',$w);
\$\endgroup\$
3
  • \$\begingroup\$ +1 Very nice. :) You could drop the trim(), actually, and in the regex you can remove the anchors and use . instead of \w. \$\endgroup\$
    – Tomalak
    Aug 3 '11 at 20:20
  • \$\begingroup\$ @Tomalak Suggested I try rewriting this solution in Perl. Including his suggestions, I got this: use List::Util 'shuffle';sub r{$_[0]=~m/(.)(.+)(.)/;$1.join('',shuffle split//,$2).$3;} That's 87 characters. Without the use line, it's 62 characters. \$\endgroup\$ Aug 3 '11 at 21:52
  • \$\begingroup\$ Can you provide a demo of this working? Because I can't... \$\endgroup\$ Sep 23 '11 at 22:33
6
\$\begingroup\$

Perl - 96 (or 71) characters 84 (or 59) characters

This is what I came up with in Perl. Went through a few different ways to do it but this seemed shortest from what I can think of so far, at 97 characters.

use List::Util 'shuffle';sub r{($b,@w)=split//,$_[0];$e=pop(@w);return$b.join('',shuffle@w).$e;}

Though, if you cut out the 'use' line (which I guess is valid, since others excluded #include lines in their C programs) I can cut it down further to 71 characters:

sub r{($b,@w)=split//,$_[0];$e=pop(@w);return$b.join('',shuffle@w).$e;}

EDIT It was suggested that I try doing this implementing @tobius' method. This way I got it down to 84 characters, or by removing the use line, 59 characters:

use List::Util 'shuffle';sub r{$_[0]=~m/(.)(.+)(.)/;$1.join'',shuffle split//,$2.$3}
\$\endgroup\$
5
  • 2
    \$\begingroup\$ shortened your Version down to 87: use List::Util 'shuffle';sub r{($b,@w)=split//,$_[0];$e=pop@w;join'',$b,(shuffle@w),$e} \$\endgroup\$
    – mbx
    Aug 3 '11 at 18:52
  • 1
    \$\begingroup\$ @sidran32 Can you implement Perl a variant of @tobius' answer, just for comparison? \$\endgroup\$
    – Tomalak
    Aug 3 '11 at 20:19
  • \$\begingroup\$ @Tomalak Sure, I'll try it out. \$\endgroup\$ Aug 3 '11 at 21:39
  • 1
    \$\begingroup\$ shortened your regex version down by 3 characters: use List::Util 'shuffle';sub r{$_[0]=~m/(.)(.+)(.)/;$1.join'',shuffle split//,$2.$3} \$\endgroup\$
    – mbx
    Aug 4 '11 at 9:26
  • \$\begingroup\$ Nice. I'm too used to using a heaping helping of parenthesis for clarity. Bad habit when golfing. :P \$\endgroup\$ Aug 4 '11 at 18:11
5
\$\begingroup\$

Ruby, 77 75 characters

def r(s);f=s.size-2;1.upto(f){|i|x=rand(f)+1;t=s[i];s[i]=s[x];s[x]=t};s;end

My Scala solution in a slightly less verbose language. I'm not a Ruby expert by any means, so there's probably room for improvement.

\$\endgroup\$
1
  • \$\begingroup\$ Wow. Works with 'I', as your scala solution. \$\endgroup\$ Aug 4 '11 at 3:35
5
\$\begingroup\$

Ruby 1.9, 77 48 46 44 chars

r=->w{w[h=1..-2]=[*w[h].chars].shuffle*"";w}

Disclaimer: I tuned this based on the highest ranked answer - noticed the exact same answer later on. You can check the history that I have kept true to my original idea but changed from ruby 1.8 to ruby 1.9 for short lambdas and shuffle.

If empty words are allowed then 56 54 chars

r=->w{w.empty?||w[h=1..-2]=[*w[h].chars].shuffle*"";w}
\$\endgroup\$
2
  • \$\begingroup\$ Nobody expects the spanish 'I'. \$\endgroup\$ Aug 4 '11 at 4:07
  • \$\begingroup\$ Attempted to handle cases with 0 or 1 letters as well \$\endgroup\$ Aug 4 '11 at 8:14
5
\$\begingroup\$

Python 3, 94 93 91 characters

Using a different technique. Might also work in Python 2.

from random import*
s=lambda x:x[0]+''.join(sample(x[1:-1],len(x)-2))+x[-1]if x[0:-1]else x

The ... if x[0:-1] else x gives x if its length is 1 (otherwise it would be duplicated). The function thereby works for strings of length 0 and 1.

The sample() is from https://stackoverflow.com/questions/2668312/shuffle-string-in-python/2668366#2668366.

Since it's one expression, we can use a lambda (eliminating return, def, and a pair of parentheses).

Edit: from random import* to save 1 character, after the other Python submission.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ I know I am so so so late here, but can x[0:-1] become x[:-1]? \$\endgroup\$
    – Adalynn
    Jul 20 '17 at 13:22
4
\$\begingroup\$

D, 62 chars

import std.random;void s(char[] s){randomShuffle(s[1..$-1]);}

okay I cheated with a normal char array instead of a real string (which is immutable char[] so no in-place shuffling)

edit with a length check it requires 14 more

import std.random;void s(char[] s){if(s.length>1)randomShuffle(s[1..$-1]);}
\$\endgroup\$
6
  • \$\begingroup\$ And it returns what for an input like 'I'? \$\endgroup\$ Aug 4 '11 at 3:23
  • \$\begingroup\$ It would be fairer (= better comparable) to return the result. \$\endgroup\$ Aug 4 '11 at 10:46
  • \$\begingroup\$ @user a range error. @ konrad that would require return s; and char[] return type 11 more chars \$\endgroup\$ Aug 4 '11 at 13:09
  • \$\begingroup\$ @ratchet Could you post the entire program, please? BTW, I don't understand why you are counting import std.random;, and not just the function. \$\endgroup\$
    – Arlen
    Aug 4 '11 at 20:11
  • \$\begingroup\$ I'm guessing you could save 1 byte by using omitting the space in char[] s (to make it char[]s), but I haven't used D in years. \$\endgroup\$
    – Tim Čas
    Dec 12 '16 at 21:42
4
\$\begingroup\$

php 5.3 (60 chars)

$r=!$w[2]?:$w[0].str_shuffle(substr($w,1,-1)).substr($w,-1);

Improved to 56 chars and no longer requires version 5.3:

$r=substr_replace($w,str_shuffle(substr($w,1,-1)),1,-1);
\$\endgroup\$
3
  • \$\begingroup\$ +1 nice alternative to the other PHP answer. Not shorter, but no regex is a plus. \$\endgroup\$
    – Tomalak
    Aug 4 '11 at 17:30
  • \$\begingroup\$ Updated with a shorter solution that doesn't require version 5.3 \$\endgroup\$
    – migimaru
    Aug 4 '11 at 18:12
  • \$\begingroup\$ The old version is not correct: returns true for short strings. \$\endgroup\$
    – Titus
    Sep 12 '16 at 19:15
4
\$\begingroup\$

Python, 86 chars

Slnicig is safe, so no bnouds ckhnceig is neeacrssy. Wkros on all leghtns.

from random import*
def f(x):x=list(x);t=x[1:-1];shuffle(t);x[1:-1]=t;return''.join(x)
\$\endgroup\$
4
\$\begingroup\$

JavaScript - 118 122 125 chars

Uses approximately the same algorithm as the OP, but with less chaining. I tried a lot of recursion, and I tried some iteration, but they all tend to get bogged down in some way or another.

function s(w){w=w.split('');var a=w.shift(),z=w.pop();return z?a+(w.sort(function(){return Math.random()-.5}).join(''))+z:a;}

Ungolfed:

function s(w)
{
    w = w.split('');
    var a = w.shift(),
        z = w.pop();
    return z?a + (w.sort(function() { return Math.random() - .5}).join('')) + z:a;
}
\$\endgroup\$
8
  • \$\begingroup\$ Does not pass the 'I'-test. \$\endgroup\$ Aug 4 '11 at 3:31
  • \$\begingroup\$ @Ryan Using return z?a+...+z:w; as an implicit length check would be in order. The silent assumption was that the function would receive only "valid" words. \$\endgroup\$
    – Tomalak
    Aug 4 '11 at 7:06
  • \$\begingroup\$ Good point, except that w has been modified, so I have to use a in the else of the ternary. Edited, and up to 122 chars. \$\endgroup\$
    – Ryan Kinal
    Aug 4 '11 at 11:57
  • \$\begingroup\$ @Ryan: I believe a would be wrong for two-letter input. :-\ Damn next time I will line out requirements more carefully. \$\endgroup\$
    – Tomalak
    Aug 4 '11 at 16:40
  • \$\begingroup\$ I don't think it would, actually. z will only be undefined if the word is one letter (or less). \$\endgroup\$
    – Ryan Kinal
    Aug 4 '11 at 17:02
3
\$\begingroup\$

Perl - 111 characters (without using any library function)

sub r{($f,@w)=split//,shift;$l=pop@w;while(@w){if(rand(9)>1){push@w,shift@w}else{push@t,pop@w}}join'',$f,@t,$l}

Usage:

$in="randomizethis";
$out = &r($in);
print "\nout: $out";
sub r{($f,@w)=split//,shift;$l=pop@w;while(@w){if(rand(9)>1){push@w,shift@w}else{push@t,pop@w}}join'',$f,@t,$l}
\$\endgroup\$
3
\$\begingroup\$

Python

It's 90 89 112 characters of python!

Edit 1: as a function this time!

(thanks gnibbler)

Edit 2: now handles short words

(thanks user unknown)

import random as r
def q(s):
 a=list(s)
 b=a[1:-1]
 r.shuffle(b)
 if len(s)<4:
  return s
 return a[0]+''.join(b)+a[-1]
\$\endgroup\$
5
  • \$\begingroup\$ even less if you follow the spec and write a function :) \$\endgroup\$
    – gnibbler
    Aug 3 '11 at 23:08
  • \$\begingroup\$ ah, pity shuffle doesn't work on strings \$\endgroup\$
    – gnibbler
    Aug 3 '11 at 23:28
  • 1
    \$\begingroup\$ The random module is like me at a nightclub... we both just sort of shuffle in place! :) \$\endgroup\$
    – Andbdrew
    Aug 3 '11 at 23:38
  • \$\begingroup\$ Doesn't work for input like 'I'; return 'II'. \$\endgroup\$ Aug 4 '11 at 3:24
  • \$\begingroup\$ thanks! it now handles short words, but is slightly longer :) \$\endgroup\$
    – Andbdrew
    Aug 4 '11 at 12:55
3
\$\begingroup\$

Scala, 135 139 142 156 characters

def r(s:String)={var(x,o,t,f)=(0,s.toArray,' ',s.size-2)
for(i<-1 to f){t=o(i)
x=util.Random.nextInt(f)+1
o(i)=o(x)
o(x)=t}
o.mkString}

-7: removed ':String' (return type can be inferred)
-7: removed 'return ' (last expression is the return value)
-3: factored s.size-2 out
-4: toCharArray -> toArray

\$\endgroup\$
2
  • \$\begingroup\$ Works with 'I' and 'Verwürfel' without fancy Python characters. :) However, my solution using 'shuffle' is a bit shorter. \$\endgroup\$ Aug 4 '11 at 3:49
  • \$\begingroup\$ @user unknown Thanks for the edits :-) \$\endgroup\$
    – Gareth
    Aug 4 '11 at 17:27
3
\$\begingroup\$

C++11: - 68 66 chars

auto f=[&](){if(s.size()>2)random_shuffle(s.begin()+1,s.end()-1);};

full program:

#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

int main(int argc, char* argv[]){

  string s = "SomestrinG";
  auto f=[&](){if(s.size()>2)random_shuffle(s.begin()+1,s.end()-1);};

  f();
  cout << s << endl;
  return 0;
}
\$\endgroup\$
2
  • \$\begingroup\$ Is hard coding in the input string legal? \$\endgroup\$ Aug 4 '11 at 21:38
  • \$\begingroup\$ @trinithis I thought we were only concerned with the function itself. The program only shows how to use the function. Regardless, not hard coding the input would not make a difference in this case; just add string s; cin >> s; \$\endgroup\$
    – Arlen
    Aug 4 '11 at 21:55
3
\$\begingroup\$

Python - 76 characters

import random as r
def f(w):m=list(w)[1:-1];r.shuffle(m);return w[0]+''.join(m)+w[-1]
\$\endgroup\$
1
3
\$\begingroup\$

R, 104 (126)

f=function(w){s=strsplit(w,"")[[1]];paste(c(s[1],sample(s[2:(length(s)-1)]),s[length(s)]),collapse="")}

Usage:

for (i in 1:10) print(f("parola"))
[1] "plraoa"
[1] "prolaa"
[1] "praola"
[1] "parloa"
[1] "plaora"
[1] "palroa"
[1] "porlaa"
[1] "ploraa"
[1] "porlaa"
[1] "ploraa"

the below function works with words with length less than 3:

f=function(w){s=strsplit(w,"")[[1]];ifelse(length(s)<3,w,paste(c(s[1],sample(s[2:(length(s)-1)]),s[length(s)]),collapse=""))}

f("pl")
[1] "pl"
f("a")
[1] "a"
\$\endgroup\$
4
  • \$\begingroup\$ Part of the task was not to move the first and last letters. \$\endgroup\$
    – Tomalak
    Mar 18 '12 at 19:15
  • \$\begingroup\$ @Tomalak fixed! \$\endgroup\$
    – Paolo
    Mar 19 '12 at 8:19
  • \$\begingroup\$ Does it work with words below the length of 3? \$\endgroup\$
    – Tomalak
    Mar 19 '12 at 8:27
  • \$\begingroup\$ @Tomalak Now it should be ok! Thanks for the corrections! \$\endgroup\$
    – Paolo
    Mar 19 '12 at 8:48
3
\$\begingroup\$

R, 95 92 91 characters

f=function(w,a=el(strsplit(w,'')),b=length(a))cat(a[1],sample(a[c(1,b)],b-2),a[b],sep="")

Makes use of R's lazy evaluation to compute a and b as function parameters, saving space with re-use later on. Also unlike other R answer this works for all words >1 char long. Example below:

> f("hippopotamus")
hpuoopaitmps

> f("dog")
dog

> f("az")
az

Edit: Replaced unlist() with [[]] Replaced [[1]] with el()

\$\endgroup\$
2
\$\begingroup\$

D: 55 characters

void f(T)(T s){if(s.length>2)randomShuffle(s[1..$-1]);};

full program:

import std.stdio, std.random, std.conv;

void f(T)(T s){if(s.length>2)randomShuffle(s[1..$-1]);};

void main(){

  char[] s = to!(char[])("SomestrinG");

  f(s);
  writeln(s);
}
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3
  • \$\begingroup\$ I think the else s part is missing? \$\endgroup\$
    – Tomalak
    Aug 5 '11 at 4:28
  • 1
    \$\begingroup\$ @Tomalak No, it's not, because there is no need for it. If the string is of length 2 or less, then we leave it alone. Also, randomShuffle() is in-place. \$\endgroup\$
    – Arlen
    Aug 5 '11 at 5:07
  • \$\begingroup\$ Wow, D being competitive. I think randomShuffle(s[1..$-1]) can be s[1..$-1].randomShuffle IIRC (unless that's in a D version older than this post) \$\endgroup\$
    – Adalynn
    Jul 20 '17 at 13:24
2
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Erlang, 188 172 132 chars

f([H|C=[_|_]])->T=[lists:last(C)],[H|s(C--T,T)];f(X)->X. s([],N)->N;s(D,N)->E=[lists:nth(random:uniform(length(D)),D)],s(D--E,E++N).

I'm still learning Erlang so any tips on making this shorter are appreciated.

full code(string_shuffle module):

-module(string_shuffle).
-export([f/1]).

f([H|C=[_|_]])->
    T=[lists:last(C)],
    [H|s(C--T,T)];f(X)->X.
f(X)->X.

s([],N)->N;
s(D,N)->
    E=[lists:nth(random:uniform(length(D)),D)],
    s(D--E,E++N).

Edit

Took the shuffle part out as a seperate function which no longer requires the head and tail of the list to be passed around.

Edit 2

Restructured to remove one of the ffunction patterns, changed the shuffle function to accept only two parameters, changed lists:delete for --[], swapped a lists:reverse call for a lists:last

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2
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PowerShell, 93

filter x{if($_.length-lt3){$_}else{$_[0,-1]-join-join($_[1..($a=$_.Length-2)]|random -c $a)}}

Look, double-jointed code!

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4
  • \$\begingroup\$ Way too long, sadly. \$\endgroup\$
    – Joey
    Aug 5 '11 at 15:56
  • \$\begingroup\$ You can't beat Golfscript anyway. ;) It's way shorter than my JS solution; Powershell is pretty expressive. \$\endgroup\$
    – Tomalak
    Aug 5 '11 at 16:47
  • \$\begingroup\$ This actually demonstrates plenty of things it doesn't handle well. Usually I aim for at least beating Python :-) \$\endgroup\$
    – Joey
    Aug 6 '11 at 0:10
  • \$\begingroup\$ saved some bytes by making a var on the first call to .Length and not declaring $a, only -5 thanks to all the brackets needed.. filter x{if(($l=$_.length)-lt3){$_}else{$_[0,-1]-join-join($_[1..($l-2)]|random -c $l)}} \$\endgroup\$
    – colsw
    Dec 6 '16 at 15:10
2
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C#, 128

static string r(string w){var t="";while(w.Length>1){int n=new Random().Next(1,w.Length-1);t+=w[n];w=w.Remove(n,1);}return w+t;}
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2
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Scala: 94

def r(w:String)=if(w.size<2)w else w(0)+util.Random.shuffle(w.tail.init.toSeq).mkString+w.last

This is a riff on "user unknowns" answer. Since a String can be implicitly cast to a Seq of chars, we can leverage Seq methods to access the middle and end of the String.

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2
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Groovy, 75

r={w->w.size()<3?w:w[0]+w[1..-2].toList().sort{Math.random()}.join()+w[-1]}

assert r('a') == 'a'
assert r('it') == 'it'
assert r('cap') == 'cap'

for(x in 1..10) {
    def w = r('Honorificabilitudinitatibus')
    println w
    assert w.size()==27 && w[0]=='H' && w[26]=='s'
}
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