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

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 2

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 3

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 4

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 5

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 6

CJam, 15 bytes

CJam is younger than this challenge, so this submission is not eligible for being accepted. This submission assumes that a "word" contains at least one character (I've requested clarification from the OP).

{(\_W>\W<mr\++}

This defines a block, the equivalent of an unnamed function in CJam.

Test it here.

If full programs (STDIN to STDOUT) are also acceptable, it can be solved in 12 bytes:

l(\_W>\W<mr\

Test it here.

Explanation

(             "Slice off first character and push it on the stack.";
 \            "Swap with remaining string.";
  _W>         "Duplicate and truncate to last character.";
     \W<      "Swap with other copy and slice off last character.";
        mr    "Shuffle.";
          \   "Swap with last character.";
           ++ "Concatenate the three parts back together.";

Note that _W>\W< could in principle be shortened to )\ (which I had originally). However, this fails when the string is empty at that point, i.e. when it originally contained less than 2 characters.

For the full program, we read the input with l first, and we can omit the ++ since the contents of the stack are printed at the end of the program automatically, back to back.

Answer 7

Ruby, 39 characters

->s{a,*b,c=s.chars;[a,*b.shuffle,c]*''}

Answer 8

Python, 102 characters

def f(x):t=list(x)[1:-1];k='';exec'k+=t.pop(id(7)%len(t));'*len(t);return[x[0],x[0]+k+x[-1]][len(x)>1]

No imports! Works for words 1 character and above. This is my first golf entry and I was inspired by BlueEyedBeast's entry from Shortest code to produce non-deterministic output for the idea of using id(Object).

---

Explanation: It makes a list of letters from the input excluding the first and last, and repeatedly pops from this list and appends to a new one until empty. The index it pops from is id(7) % len(list we're popping from). Since id(7) is the memory address of the object 7, it is essentially random. So now we have a list of randomly scrambled letters from the center of the original input. All we do now is append the first and last letters of the original output accordingly and we got the output we want: (first letter)+(scrambled middle)+(last letter).

Answer 9

Python 3, 80 chars

This is an improvement on Ashley Grenon's answer. By removing variable assignments and using random.sample instead of random.shuffle, so you don't have to explicitly call list(), I saved 4 characters. This one still fails on strings less than 2 characters.

import random as r
def f(w):return w[0]+''.join(r.sample(w[1:-1],len(w)-2))+w[-1]

Answer 10

Perl, 41 bytes

40 bytes code + 1 for -n.

$\=chop;s/.//;print"$&",sort{rand 2}/./g

Requires input to be supplied without a newline (echo -n ...) as we use chop to move the last char into $\ (which is appended after all arguments to print) and we sort, applying 0 or 1 (rand 2) to each character. This appears to result in all permutations given enough time. The $& has to be in quotes, as it changes after the /./g is executed, interpolating into a string works around this. Also print has to be used instead of say to utilise $\ :(.

Usage

echo -n 'stringified' | perl -ne '$\=chop;s/.//;print"$&",sort{rand 2}/./g'
sifirgtined

for i in `seq 1 20`; do echo -n 'stringified' | perl -ne '$\=chop;s/.//;print"$&",sort{rand 2}/./g'; echo; done 
segfiiintrd
sitnrigfied
stringified
srtinigfied
sginitrefid
strnigieifd
seriftniigd
sifertniigd
siirtngfied
sintrgieifd
snitrigefid
sfgieinirtd
sigitrnifed
sefiirtingd
sirtnigifed
sifgenritid
srtingiifed
siignrteifd
sefiigtrind
sifneirtigd

Answer 11

R, 93 79 bytes

function(x,l=nchar(x))"if"(l<4,x,intToUtf8(utf8ToInt(x)[c(1,sample(l-2)+1,l)]))

Try it online!

Wroks for any ipnut szie dwon to zero, by frsit ckciheng the lgnteh of the sintrg and rneturing it if nchar(x)<4

-14 bytes thanks to Giuseppe!

Answer 12

Powershell, 81 bytes

Inspired by Joey: I like her double join.

$args|%{($_,($_[0,-1]-join-join($_[1..($a=$_.Length-2)]|random -c 2e9)))[$a-ge2]}

Notable changes:

1. if else replaced with a Fake ternary operator.
2. $_.Length is written once.
3. Other golf to compact the script
4. The script gets Random for the word length >= 4 only.
5. random -c $a replaced with random -c 2e9. It need to avoid an exception with word length <= 2. I think the length = 2000000000 is enough for a word.

Test script:

$f = {

$args|%{($_,($_[0,-1]-join-join($_[1..($a=$_.Length-2)]|random -c 2e9)))[$a-ge2]}

}

&$f ""
&$f a
&$f ab
&$f abc
&$f abcd
&$f randomize
&$f reading
&$f Hello
&$f world

Output:

a
ab
abc
acbd
rmaniodze
rniedag
Hlelo
wrlod

Answer 13

Vyxal, 5 bytes

ÞẇṖ℅j

Try it Online!

Þẇ    # Push (first+last, rest)
  Ṗ℅  # Shuffle (rest)
    j # Join (first+last) by (rest)

Answer 14

Scala: 112 123

def v(s:String)=if(s.size<4)s else s(0)+util.Random.shuffle(s.substring(1,s.size-1).toList).mkString+s(s.size-1)

Incorporated hints from the comments (length->size (I need this hint every time), mkString without ("")) and size-1, not -2). Thanks.

Answer 15

Lua, 173 chars

Does what it needs to do.

function(a)return(a:gsub('(%w)(%w*)(%w)',function(a,b,c)t={}for l in b:gmatch'.'do t[#t+1]=l end while #t>0 do i=math.random(#t)a=a..table.remove(t,i)end return a..c end))end

Answer 16

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.

Answer 17

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 18

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 19

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 20

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 21

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 22

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 23

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 24

TinyMUSH, 102

&_ me=left(%0,1)[ifelse(eq(strlen(%0),1),,scramble(mid(%0,1,sub(strlen(%0),2)))[right(%0,1)])]
\u(_,X)

Replace "X" with the input word. The user-defined _ attribute uses the built-in scramble() function to scramble letters between the first and last letters.

Answer 25

Clip, 24

?<lxWx],*R>%Ox{(x)x`(x)x

Here's an effective translation in pseudocode:

x = input by stdin
if len(x) < 2:
  output x
else:
  y = rotate_right(x, 1)
  y.remove_first_occurrence(first(x), last(x))
  # y is now the characters between the first and last of the input.
  shuffle(y)
  output place_at_end(last(x), place_at_beginning(first(x), y))

Answer 26

Rebol - 63

f: func[s][e: take/last s ajoin trim reduce[take s random s e]]

Ungolfed:

f: func [s] [
    e: take/last s 
    ajoin trim reduce [take s random s e]
]

Usage example:

>> f "1"
== "1"

>> f "12"
== "12"

>> f "123"
== "123"

>> f "1234"
== "1324"

NB. An alternative (but unfortunately not shorter) version would be:

f: func [s] [
    change/part next s random copy/part next s back tail s back tail s
    s
]

With golfing and minor shavings this would come out at 73 chars.

Answer 27

MATLAB, 48 bytes

function t=f(t)
t(2:end-1)=t(randperm(end-2)+1);

Too bad the spec says 'function', using input() would be 43 bytes. Another version, that will fail slightly on a one-length word (a becomes aa), is 34 bytes:

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

Answer 28

Pyth - 11 bytes

Pyth is younger than this challenge so it doesn't count.

pez+hz.SPtz

Gets middle, shuffles it, then adds the rest. Takes I/O from stdout/stdin.

pez+hz      Print First letter of input, something else, and last letter of input
 .S         Shuffle
  P         All but the last
   t        All but the first
    z       The input

Answer 29

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 30

Coffee-script: 76 91 Bytes

f=(a)->a.length>1&&a[0]+(a.split('')[1..-2].sort ()->.5-Math.random()).join('')+a[-1..]||a

Wow, I like this language already.