Randomly Misspell a Word

Question

Inspired by this CR question ^{(please don't kill me for browsing CR)}

Spec

The probabilities of misspelling a word are:

• 1/3 of the time don't change the output
• 1/3 of the time remove a random character
• 1/3 of the time duplicate a random character

The chance for removing/duplicating a given character in the input should be the same for all chars.

If two consecutive characters are the same (case-sensitive), the probability of one of them being modified should be the same as if they are one character. I.e. the outputs for AA (which are AA or A or AAA) should all have the same probability.

---

The input will only contain of letters for simplicity.

Examples

First line is input, following lines are all possible misspellings. Each line should have the same probability of being output, the input is excluded in the examples but it should still have a 1/3 probability of output.

foo

fo
oo
ffoo
fooo

PPCG

PPC
PPG
PCG
PPPCG
PPCCG
PPCGG

Answer 1

Pip, 38 27 bytes

a@:`(.)\1*`YRR#aa@y@0X:RR3a

This was fun--got to use Pip's regex and mutable string capabilities, which I hadn't pulled out in a while. Takes input via command-line argument.

Explanation:

a@:`(.)\1*`                  Split a into runs of identical chars using regex match
           YRR#a             Yank randrange(len(a)) into y (randomly choosing one such run)
                a@y@0        Take the first character of that run
                     X:RR3   Modify in place, string-multiplying by randrange(3):
                               If RR3 is 0, character is deleted
                               If RR3 is 1, no change
                               If RR3 is 2, character is duplicated
                          a  Output the modified a

Try it online!

Answer 2

Ruby, 64 55 + 1 (p flag) = 56 bytes

Input is a line of STDIN piped in without trailing newline.

a=[]
gsub(/(.)\1*/){a<<$&}
a.sample[-1]*=rand 3
$_=a*''

Answer 3

CJam (21 bytes)

re`_,mr_2$=3mr(a.+te~

Online demo

Dissection

r     e# Read a line of input from stdin
e`    e# Run-length encode it
_,mr  e# Select a random index in the RLE array
_     e# Hang on to a copy of that index
2$=   e# Copy the [run-length char] pair from that index
3mr(  e# Select a uniformly random integer from {-1, 0, 1}
a.+   e# Add it to the run-length
t     e# Replace the pair at that index
e~    e# Run-length decode

Answer 4

JavaScript (ES6), 107

w=>(r=x=>Math.random()*x|0,a=w.match(/(.)\1*/g),a[p=r(a.length)]=[b=a[p],b+b[0],b.slice(1)][r(3)],a.join``)

Less golfed

w=>(
  a = w.match(/(.)\1*/g),
  r = x => Math.random()*x | 0,
  p = r(a.length),
  b = a[p],
  a[p] = [b, b+b[0], b.slice(1)][r(3)],
  a.join``
)

Test

f=w=>(r=x=>Math.random()*x|0,a=w.match(/(.)\1*/g),a[p=r(a.length)]=[b=a[p],b+b[0],b.slice(1)][r(3)],a.join``)

function update() { 
  O.innerHTML = Array(99)
  .fill(I.value)
  .map(x=>(
    r=f(x),
    r==x?r:r.length<x.length?'<b>'+r+'</b>':'<i>'+r+'</i>'
  
    )
  ).join` `
}

update()

#O { width:90%; overflow: auto; white-space: pre-wrap}

<input id=I oninput='update()' value='trolley'><pre id=O></pre>

Answer 5

Java 7, 189 180 178 bytes

import java.util.*;String c(String i){Random r=new Random();int x=r.nextInt(2),j=r.nextInt(i.length());return x<1?i:i.substring(0,j-(x%2^1))+(x<2?i.charAt(j):"")+i.substring(j);}

Ungolfed & test cases:

Try it here.

import java.util.*;
class M{
  static String c(String i){
    Random r = new Random();
    int x = r.nextInt(2),
        j = r.nextInt(i.length());
    return x < 1
            ? i
            : i.substring(0, j - (x%2 ^ 1)) + (x<2?i.charAt(j):"") + i.substring(j);
  }

  public static void main(String[] a){
    for(int i = 0; i < 5; i++){
      System.out.println(c("foo"));
    }
    System.out.println();
    for(int i = 0; i < 5; i++){
      System.out.println(c("PPCG"));
    }
  }
}

Possible output:

foo
fooo
foo
foo
ffoo

PPCCG
PPCG
PPCCG
PPPCG
PPCG

Answer 6

Python 2, 134 bytes

from random import*
def f(s):
 p=c=0;M,L=[],list(s)
 for t in L:
  if t!=p:M+=c,;p=t
  c+=1
 L[choice(M)]*=randint(0,2);return''.join(L)

White spaces in for loop are tabs.

Try it on Ideone

Answer 7

Pyth - 17 bytes

This one actually handles the special cases with consecutive char correctly.

 XZOKrz8Or_1 2r9K

Test Suite.

Answer 8

APL, 21

{⍵/⍨3|1+(?3)×(⍳=?)⍴⍵}

This starts by creating a vector of zeros with a 1 in random position. Then multiplies it by a random number between 1 and 3. +1 and mod 3 obtains a vector with all 1s and one random positioned 0,1 or 2.

Finally, ⍵/⍨ says that each letter should be written n times, where n are the numbers in the vector.

Try it on tryapl.org

Answer 9

Python 2, 123 bytes

from random import*
R=randint
s=input()
m=n=R(0,len(s)-1)
c=R(0,2)
m=[m,1+[-len(s),m][m>0]][c==1]
n+=c==2
print s[:m]+s[n:]

Answer 10

JavaScript (ES6), 103

w=>{w=w.split(''),r=Math.random,x=r(),i=r()*w.length|0;w.splice(i,x<.6,x>.3?w[i]:'');alert(w.join(''))}

Answer 11

APL, 27 bytes

{⍵/⍨(?3)⌷1⍪2 0∘.|1+(⍳=?)⍴⍵}

Explanation:

                        ⍴⍵  ⍝ length of ⍵
                   (⍳=?)    ⍝ bit vector with one random bit on 
                 1+         ⍝ add one to each value (duplicate one character)
           2 0∘.|           ⍝ mod by 2 and 0 (remove instead of duplicate)
         1⍪                 ⍝ add a line of just ones (do nothing)
    (?3)⌷                   ⍝ select a random line from that matrix
 ⍵/⍨                        ⍝ replicate characters in ⍵

Test:

      {⍵/⍨(?3)⌷1⍪2 0∘.|1+(⍳=?)⍴⍵} ¨ 10/⊂'foo'
 fooo  foo  oo  foo  fo  fo  oo  fo  oo  fooo