How to randomize letters in a word

Question

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.

Answer 1

Mathematica 98 chars

Not the most economical code but fun to write.

s holds the input sentence (as a string).

Row[StringJoin @@@ (Characters@# /. {f_, m___, e_} :>  
   Flatten@{f, RandomSample@{m}, e} & /@ StringSplit@s), " "]

For example, when `s = "This sentence is fairly easy to read.",

Answer 2

J, 25

(]/:0,((2-~#)?1-~#),1-~#)

Example:

      (]/:0,((2-~#)?1-~#),1-~#) 'order'
oedrr
      (]/:0,((2-~#)?1-~#),1-~#) 'order'
oerdr
      (]/:0,((2-~#)?1-~#),1-~#) 'order'
oderr
      (]/:0,((2-~#)?1-~#),1-~#) 'order'
odrer
      (]/:0,((2-~#)?1-~#),1-~#) 'order'
oerdr
      (]/:0,((2-~#)?1-~#),1-~#) 'order'
oerdr

Answer 3

APL (Dyalog), 34 Charachters

Still trying to golf it a bit more, new to APL. Tips appreciated.

y←(⍴x←⍞)-2⋄x[1],x[1+⍳y][y?y],x[⍴x]

Here is an attempt to explain it, I also simplified it a bit (no charachter improvement, though)

⋄ is a statement separator, think of it as a new line.

That leaves us with 2 statements. y←(⍴x←⍞)-2 and x[1],x[1+⍳y][y?y],x[⍴x]

APL works from right to left in statements, but follows parentesis still, so (⍴x←⍞) is executed first. ⍞ takes charachter input. ← assigns that to x and ⍴ gives the length of x. Then the -2 is executed, which subtracts 2 from the length of x. Finally, the length-2 is assigned to y and we move on to the next statement.

x[⍴x] takes the last character of x, think of it as x[x.length] (using the length as the index of the last character).

, is catenate.

So we concatenate the last character of x with x[1+⍳y][y?y] which takes the middle indices of x using 2+⍳y and applies a randomization using [y?y].

⍳y generates 1 2 3 ... y and 1+ turns this into 2 3 4 ... y+1 which are the middle indices of x, for example, this returns bcdef from abcdefg.

[y?y] "deals" y values from 1 to y.

So, x[1+⍳y][y?y] grabs the middle of the word and randomizes it.

Finally, we concatenate the first charachter of x using x[1], to the rest of the string, and that is the output of the program.

Hopefully that was understandable...

Answer 4

C# w/Linq - 152 non-whitespace chars. It's terrible compared to other languages on char count, but elegant:

public string Shuf(string i)
{
    return new String(i.Take(1)
        .Concat(i.Skip(1).Take(i.Length-2).OrderBy(x=>Guid.NewGuid()))
        .Concat(new[]{i.Last()})
        .ToArray());
}

Answer 5

I know this is an older puzzle, but I wanted to add my two cents from VBA: 128 177 151:

Sub o(s)
n=Left(s,1)
p=Len(s)-1
s=Right(s,p)
For i=1 To p-1
p=p-1
r=Int(p*Rnd()+1)
n=n & Mid(s,r,1)
s=Left(s,r-1) & Right(s,p-r+1)
Next
s=n & s
End Sub

Example Usage:

a = "According"
o a ' a is assigned by the passing of the reference to the 'o' sub.
a = "to"
o a ' see above...
a = "some"
o a
a = "controversial"
o a
a = "story"
o a

I was actually happily surprised to see it fare as well as it did against some of the other, more typical CG languages. After improving, this is still not the shortest, but I was happy to do it.

Answer 6

MATLAB, 65 47 characters

r=@(s)s([3-min(end,2):1 randperm(end-2)+1 end])

Works for strings of positive length, but fails on the empty string. Improved by using end as in jazzkingrt's solution.

This use of end can further improve jazzkingrt's solution by substituting end for length(x). However, that solution doesn't handle strings of length one correctly. (I'm not allowed to comment, so I write here instead.)

Answer 7

MATLAB (46 Characters)

f=@(x)[x(1) x(randperm(length(x)-2)+1) x(end)]

Sample Usage:

>> f('elephant')

ans =

eplhnaet

>> f('imawesomebutyousuck')

ans =

iouesutuoeswbamycmk

Works with words size two or greater.

Answer 8

Q (42 Characters)

{(x[0]),((neg count 1_-1_x)?1_-1_x),-1#x}

Sample Usage:

q){(x[0]),((neg count 1_-1_x)?1_-1_x),-1#x} "elephant"
"eeahlnpt"
q){(x[0]),((neg count 1_-1_x)?1_-1_x),-1#x} "ant"
"ant"

Answer 9

Q, 38 48

{((:)x),((-1(#:)a)?a:-1_1_x),last x}

Had to change it for words <=3 letters

{$[3<(#)x;((*:)x),((-1*(#:)a)?a:-1_1_x),-1#x;x]}

Answer 10

JavaScript (132 characters):

function r(w){return w.substr(0,1)+w.substr(1,w.length-2).split('').sort(function(){return Math.random()-.5}).join('')+w.substr(-1)}

You don't need to go through point by point. If provided a function which returns something where Math.round will return a random 0,1,-1, that will be sufficient (because of how sort works).

Answer 11

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"

Answer 12

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'
}

Answer 13

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.

Answer 14

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;}

Answer 15

Perl - 76 chars

Pure Perl implementation that requires no external modules. Also correctly handles one-letter, two-letter word edge cases.

sub r{@a=split//,pop;@a>1?join'',@a[0,(sort{rand()<=>rand}1..$#a-1),$#a]:@a}

---

Usage

say r('stringified') for 1 .. 20;

# Example output

srfgiietnid
sgfnieiitrd
seiifgrtnid
siigrfeintd
sgiirntiefd
siigerniftd
sifiitgnred
sftrneiiigd
sfieirtngid
sfeitirgnid
sfiertnigid
sigteifrind
sgieftirnid
sieitfrnigd
stnigirfeid
sfietrniigd
siigfrenitd
stefrniiigd
setrfiniigd
sifgtiernid

---

N.B. _{I'm not entirely convinced that the letter randomization still renders the word readable in the example above}

Answer 16

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]

Answer 17

Haskell, 4 characters

The function trinithis proposed actually matches the specification:

s=id

It returns the string unchanged, thus keeping the first and last characters in place and doing a permutation of all the other characters.

If someone is dissatisfied with the probability distribution of the permutations, here is a solution yielding a better distribution. It is obviously much more complex:

Haskell, 110 120 characters

import Random
s(a:t@(b:c:d))=do{i<-randomRIO(0,length d);fmap(([a,t!!i]++).tail)$s$a:take i t++drop(i+1)t}
s l=return l

An example of a program using this function:

main = getLine >>= s >>= putStrLn

Answer 18

Python, 86 characters

import random as r
def f(w):t=list(w[1:-1]);r.shuffle(t);return w[0]+''.join(t)+w[-1]

And here's an example of using it:

for x in ["ashley", "awesome", "apples"]:
    print f(x)

This is my first code golf exerise. After solving the problem I decided to look at the answers and it's no surprise mine answer isn't unique. This was fun though :o)

I did make one change after looking at the other responses and it was changing my import statement to use an alias. Great idea. ;o)

Answer 19

PowerShell, 93

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

Look, double-jointed code!

Answer 20

Ruby 1.9, 43 characters

r=w[0]+[*w[1..-2].chars].shuffle.join+w[-1]

Does not yet work for 1 character length Strings (duplicates that character), and fails for empty String.

Answer 21

python, 87 79 75 93 92 chars (handling strings of length 0,1)

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.

Answer 22

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 http://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.

Answer 23

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

Answer 24

Javascript, 202 characters

function r(s){var s=s.split(""),h=[-1],n=s.length-1,t,i;h[-1]=s[0];for(i=0;++i<n;){do{t=Math.random()}while(t in h);h[t]=s[i];h[i]=t;}h.sort();for(i=0;i<n;++i)h[i]=h[h[i]];return n?h.join("")+s[n]:s[0]}

This solution has an unbiased distribution.

Algorithm:

Split input string into an array s. Consider another array h that is doubly used as a dictionary. For each letter in s at index 0 < i < s.length-1, assign a unique random number to h at i. Also map the random number in the range [0, 1) in h to the letter. The first and last letters are handled specially. Before assigning the random numbers as described above, do the analogous thing for the first letter, but hard code the number as -1 (guaranteed to be less than the smallest random number generated, which can be 0). Ignore the last letter for now. Sort h. Map h's random value to the corresponding letter. Join the array into a string and tack on the last letter. Special case for 1 character input, where we return the first character (we still crunch h because the logic is shorter that way).

Answer 25

Java, 194 charcters

String r(String s){String[]c=s.split("");if(c.length>2)java.util.Collections.shuffle(java.util.Arrays.asList(c).subList(2,c.length-1));return(""+java.util.Arrays.asList(c)).replaceAll("\\W","");}

If you can assume java.util.* is imported, then you can shave off a fair amount of characters. Might be able to squeeze in a List variable to save a few more. If the class the r function is implemented in derives from java.util.Arrays, then even more characters can be saved.

Answer 26

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;
}

Answer 27

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);
}

Answer 28

FSharp, ~207 177

Count includes whitespace since its significant for f#

let f w=
 let r=System.Random()
 let e = Seq.length w
 (w|>Seq.mapi(fun i l->if i=0||i=e-1 then i,l else r.Next(1,e-1),l)|>Seq.sortBy fst|>Seq.map snd|>List.ofSeq).ToString()

Run with:

printfn "%s" (f "apples")

Answer 29

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)

Answer 30

Haskell, 195

import Random
import Data.List
r l=fmap(f l.g l)newStdGen
f l n=map snd$head$filter(\l->map fst l==n)$permutations$zip[0..]l
g l=take(n+2).(0:).(++[n+1]).take n.nub.randomRs(1,n)where n=length l-2

Hacked this into existence while Rotsor was making his post. This will cover all possible permutations, and I believe it does so uniformly.

The function r is the function that the user uses.

The algorithm is fairly straightforward. It generates a random list of unique numbers from 1 to n-2. Then it tacks on 0 to the front and n-1 to the end of the list. The take(n+2) is there to handle the case where the input is a single character (or an empty string for that matter). Then it searches for the corresponding permutation and returns that.

The list is generated by generating an infinite list of random numbers from 1 to n-2. Then it picks out the first occurance of each number in that range.