[Question]
  [
Remember those brute-force programs to crack password that show every combination they are trying? More precisely, at one point, the *n* first characters are fixed (they have been guessed successfully), and every possible character for the remaining ones is being tested. You've probably seen some in movies, or in some software written by people that like fancy interfaces.  
Sorry to disappoint, but we won't write a program to crack password, only one to reproduce the nice output.
### Challenge
Given a string containing printable ascii characters but no newlines (ascii code 32 to 126 or matching the regex `^[ -~]{2,}$`), print an output following this rules:
* At time `t=n seconds`, the `n` first characters printed are the `n` first characters of the input string.
* After the `n` fixed characters, you should append a string formed random character (chosen uniformly pseudo-randomly from the unicode range    to `~` (code 32 to 126)) to form a string of the length of the initial one.
* You should output *at least* (more on that later) 20 lines every second: every one of them will have the same `n` first characters, but a different random end.
It's probably not very clear yet what you are supposed to do, so lets go through an example:
### Example
*I'll print only 5 different lines for every seconds instead of the 20 minimum just to make it more readable.* 
Consider the input `abcde`.  
During the first second, a valid output can be something like (completely random): 
```
dGuT4
S!jkN
"gQ>[
TU3! 
*fAjV
```
Then, `t=1`, the first character of every following string will be `a` (the first character of the input):
```
a);jD
aHv^p
aqw5*
a|.?:
a{gbK
```
Now, `t=2`, the first two characters will be `ab`:
```
abTJ"
ab\ e
ab3T#
abYWS
ab"#<
```
Now, `t=3`, the first three characters will be `abc` :
```
abcvW
abc3G
abc(g
abc{@
abc@4
```
Now, `t=4`, the first four characters will be `abcd` :
```
abcdD
abcdv
abcdj
abcd$
abcd6
```
Finally, `t=5`, we print the input (only once):
```
abcde
```
### A few precisions
* You shouldn't bother too much with your language precision toward the seconds (ie. If your algorithm is correct but your system/language lacks precision then it's fine).
* The first second can be shorter than one second (That is, if you launch you program during a second, the first second can be just the remaining time until the end of the current second). Or put differently, you don't have to wait for the start of a new second to start printing the outputs.
* *At least 20 lines per second*: The more natural way would be an infinite loop with a special behaviour one every second (or a timeout, or whatever), so that will result in probably a few thousand lines per second (and that's perfectly fine!). But if you have another idea, feel free to use it as long as you print at least 20 lines per second.
* The input will always be more than 2 characters long.
* You can consider that the input won't be more that 30 characters long if it helps. (But if it works for longer ones, it's for the best)
* The input format should be the most natural representation of a string in your language.
* You are allowed to print a trailing newline.
### Code example
If you still don't understand exactly what you have to do, you can run the following code in a linux terminal to see:
```
perl -F -aplE 'map{$t=time;print$s,map{chr 32+rand 94}@F until$t-time;$s.=shift@F}@F' <<< "Cracking in progress\!"
```
### Winning criterion
This is [code-golf](/questions/tagged/code-golf "show questions tagged 'code-golf'"), so shortest code in byte wins!
---
*Thanks to Laikoni and Flp.Tkc for their suggestions and improvement in the sandbox.*
  
[Answer]
## HTML/JavaScript, ~~170~~ ~~168~~ 167 bytes
```
setInterval('o.textContent=i.value.replace(/./g,(c,i)=>new Date-d>++i*1e3?c:String.fromCharCode(Math.random()*95+32))',d=50)
```
```
<input id=i oninput=d=Date.now()><pre id=o>
```
Edit: Saved 2 bytes thanks to @ETHproductions. Saved 1 byte thanks to @jrich.
[Answer]
# Node, ~~145~~ 142 bytes
```
for(s=process.argv[2],d=new Date;s[a=(new Date-d)/1e3|0]+console.log(s.replace(/./g,(c,i)=>i<a?c:String.fromCharCode(32+Math.random()*95))););
```
This seems a little long, and there's probably a little room for golfing. Note that the semicolon at the end is required; without it the program throws a syntax error because the `for` statement has no body.
Outputs way more than 20 lines per second; a little birdie told me that it's roughly 12 thousand. Here's how it looks in the ConEmu terminal emulator on my computer (recorded at 30 fps):
[![enter image description here](https://i.stack.imgur.com/4MSm0.gif)](https://i.stack.imgur.com/4MSm0.gif)
[Answer]
# [05AB1E](http://github.com/Adriandmen/05AB1E), 26 bytes
I post this as a different answer with respect to the other 05AB1E answer since the approach is different
```
.põ¸ì¨vT·FyžQ.r¹gyg-£«}}¹»
.p                         Generate ordered prefix of input (e.g., ["a", "ab", "abc", "abcd", "abcde"] for "abcde")
  õ¸ì                      Prepend an empty string (e.g., result is ["", "a", "ab", ...])
     ¨                     Strip the last element (the same string as the input)
      v                    For each string in the array
       T·F                 For N in range(20)
          y                Push the current string
           žQ.r            Push all printable characters, shuffled
               ¹gyg-       Take the difference between the length of the input and the length of the current string -> x
                    £      Take the x first characters from the shuffled printable characters
                     «     Yield currentString + shuffledCharacters
                      }    End inner for
                       }   End outer for
                        ¹  Push input (last iteration)
                         » Join everything with newlines and implicitly display
```
[Try it online!](http://05ab1e.tryitonline.net/#code=LnDDtcK4w6zCqHZUwrdGecW-US5ywrlneWctwqPCq319wrnCuw&input=dGhpc3RleHRpc3dheXRvb2xvbmc)
[Answer]
## BASH, 99 93 92 91 88 bytes
with `tr` + `head` + `urandom`
```
while ((${#1}-n));do
echo "${1::n=SECONDS}`tr -dc \ -~</dev/ur*|head -c$[${#1}-n]`"
done
```
(thx. to @manatwork )
[Answer]
# 05AB1E, 30 bytes
```
gFžcUNV[¹Y£žQ.r¹gY-£J,XžcÊ#}¹,
```
Uses the **CP-1252** encoding. [Try it online!](http://05ab1e.tryitonline.net/#code=Z0bFvmNVTlZbwrlZwqPFvlEucsK5Z1ktwqNKLFjFvmPDiiN9wrks&input=YWJjZGU)
Please help me golf this down.
[Answer]
# C, ~~182~~ ~~176~~ ~~128~~ ~~126~~ 125 bytes
Golfed:
```
i;s;n;x;g(char*c){time(&s);while(c[++x]);do{n=time(0)-s;for(i=0;i<x;i++)putchar(i<n?c[i]:32+rand()%95);puts("");}while(n<x);}
```
Ungolfed:
```
#include "stdio.h"
#include "stdlib.h"
#include "time.h"
int i,s,n,x;
void g(char* c) {
  time(&s); //Get the initial time
  while(c[++x]); // x = strlen(c) (happy about this one)
  do {
    n = time(0) - s; //seconds since beginning
    for(i = 0; i < x; i++)
      //after each second, print another char of the password
      putchar(i < n ? c[i] : 32 + rand() % 95);
    puts("");
  } while(n < x); //while we haven't printed the whole word
}
```
~~I've heard that it's possible to drop some standard `#include`s, but I couldn't get it to work on the MingW GCC compiler I just downloaded. Also couldn't figure out how to `#define b #include` without using more space than it was worth.~~
I'm just an idiot, it works fine without them.
[Answer]
# Java 7, ~~271~~ ~~265~~ 207 bytes
```
void c(String s)throws Exception{for(int i=0,j,l=s.length();i<=l*20;i++){String r=s.substring(0,i/20);Thread.sleep(45);for(;j++<l;r+=(char)(32+Math.random()*95);System.out.println(r);if(s.equals(r))return;}}
```
-58 bytes saved thanks to *@OliverGrégoire*. ([Don't forget to upvote his even shorter Java 8 answer.](https://codegolf.stackexchange.com/a/100772/52210))
**Ungolfed:**
```
void c(String s) throws Exception{
  for(int i = 0, j, l = s.length(); i <= l*20; i++){
    String r = s.substring(0, i/20);
    Thread.sleep(45);
    for( ; j++ < l; r += (char)(32+Math.random()*95));
    System.out.println(r);
    if(s.equals(r)){
      return;
    }
  }
}
```
**Input:** `abcde`  
**Output:**
[![enter image description here](https://i.stack.imgur.com/33ooh.gif)](https://i.stack.imgur.com/33ooh.gif)
[Answer]
# WinDbg, ~~400~~ 391 bytes
```
.for(r$t1=@$t0;by(@$t1);r$t1=@$t1+1){};m@$t0 L@$t1-@$t0+1 @$t1+1;r$t4=2*@$t1+2-@$t0;r$t8=@$t4+f;r$t3=0;.for(r$t2=0;@$t2<@$t1-@$t0;da@$t0){.for(r$t7=@$t0+@$t2;by(@$t7);r$t7=@$t7+1;r$t8=@$t8+1){eb@$t7 by(@$t8)%5e+20};r$t9=0;.foreach(p {.echotime}){.if7==@$t9{ea@$t4"p";.if1>@$t3{r$t3=by(@$t4+7)}};r$t9=@$t9+1};j@$t3!=by(@$t4+7)'m@$t0+@$t4-@$t1+@$t2-1 L1 @$t0+@$t2;r$t2=@$t2+1;r$t3=by(@$t4+7)'}
```
*-9 bytes by simplifying some math*
This definitely does not seem to be the kind of thing WinDbg is intended to do. ;)
Input is taken by entering an ascii string at a memory location and setting that address to the pseudo-register `$t0`. Eg:
```
r$t0 = 2000000
eza @$t0 "abcde"
```
The prng I'm using is whatever the contents in memory, some bytes past the input string. Chrome.exe appears to fill the memory space after `0x2000000` with random-looking-enough bytes so I used a dump of chrome.exe. Unknown if this is uniform, but looks random-enough to me.
How it works:
```
.for(r$t1=@$t0; by(@$t1); r$t1=@$t1+1){};         * From $t0, increment $t1 until the byte
                                                  * at $t1 is 0 to find length of input
m@$t0 L@$t1-@$t0+1 @$t1+1;                        * Duplicate input (memory 
                                                  * becomes: "input\0input\0")
r$t4=2*@$t1+2-@$t0;                               * Set $4 to the byte after \0 of the 
                                                  * duplicated input
r$t8=@$t4+f;                                      * Set $t8 to $t4+15, this is the prng
r$t3=0;                                           * Init $t3=0, this will hold the time
.for(r$t2=0; @$t2<@$t1-@$t0; da@$t0){             * For $t2=0, loop until it's input length,
                                                  * printing the string at $t0 after each
                                                  * loop. $t0 is where the password crack
                                                  * progress is written.
    .for(r$t7=@$t0+@$t2; by(@$t7); r$t7=@$t7+1;   * Loop over each uncracked char
                                   r$t8=@$t8+1){  * also incrementing prng ($t8)
        eb@$t7 by(@$t8)%5e+20                     * Write a visible ascii char onto the
                                                  * uncracked char position based on the 
                                                  * current byte of prng%0x5e+0x20 (prng%126+32)
    };
    r$t9=0;                                       * Set $t9=0 for updating current time
    .foreach(p {.echotime}){                      * For each (string) word in a statement
                                                  * like "Debugger (not debuggee) time: Mon 
                                                  * Nov 21 18:23:08.433 2016 (UTC - 8:00)"
        .if7==@$t9{                               * If the 7th word, ie- the current time
            ea@$t4"p";                            * Write the time at $t4
            .if1>@$t3{                            * If $t3 has not been set yet
                r$t3=by(@$t4+7)                   * ...save the current second in $t3
            }
        };
        r$t9=@$t9+1                               * Increment $t9 until it's 7
    };
    j@$t3!=by(@$t4+7)'                            * If the current second has changed
        m@$t0+@$t4-@$t1+@$t2-1 L1 @$t0+@$t2;      * Copy the cracked char from dupe input
        r$t2=@$t2+1;                              * Increment $t2 (loop ends when this is input length)
        r$t3=by(@$t4+7)                           * Save the new current second
    '
}                                                 * Final crack is printed by for loop
```
Note: Some bytes could be golfed by using `j` instead of the `.if`'s, but that causes it to run too slowly on my machine so it doesn't output at least 20 lines per second, so not saving those bytes.
Sample Output: <http://pastebin.com/H4H74sAx>
[Answer]
## R, 138 bytes
```
z=Sys.time;n=nchar(x<-scan(,""));s=z();t=0;while(t<=n){t=t+z()-s;cat(substr(x,1,f<-floor(t)),intToUtf8(sample(32:126,n-f,T)),"\n",sep="")}
```
Reads input from stdin.
Counted approximately 61 lines on my machine between each additional letter in the "password".
[Answer]
# Bash, ~~247~~ ~~245~~ ~~212~~ 207 bytes
```
R()(echo $SECONDS);w=`R`;until [ "$a" = "$1" ];do for i in `seq 1 $[${#1}-${#a}]`;{ a+=`printf "\x$(printf %x $[$RANDOM%127+32])"`;};echo -e "$a\r";a=${1:0:q};((`R`-w>0))&&{ w=`R`;((q++));}||:;done;echo "$a"
```
*Thanks a lot Bash for being so whitespace sensitive...*
Anyways, output is given in real time on separate lines. Save as a `.sh` script and invoke with:
`bash <File Name>.sh <Input>`
For example, `bash Cracking_In_Progress.sh okayerty` results in the following output, recorded at 30 frames per second:
[![Example Output](https://i.stack.imgur.com/797Md.gif)](https://i.stack.imgur.com/797Md.gif)
[Answer]
# Haskell (GHC), 202 bytes
```
import System.Random
import Control.Concurrent
f s|l<-length s=mapM_(\n->putStr('\r':take n s)>>mapM(\_->toEnum<$>randomRIO(32,126))[1..l-n]>>=putStr>>threadDelay 50000)$[n|n<-[0..l-1],f<-[1..20]]++[l]
```
-5 bytes without fancy carriage return action
[![enter image description here](https://i.stack.imgur.com/3SZjR.gif)](https://i.stack.imgur.com/3SZjR.gif)
[Answer]
# [MATL](http://github.com/lmendo/MATL), 26 bytes
```
`GZ`:)' ~'olGn4Mk-I$YrhD7M
```
[Try it online!](http://matl.tryitonline.net/#code=YEdaYDopJyB-J29sR240TWstSSRZcmhEN00&input=J2FiY2RlJw)
Below is the real-time output from the offline compiler. Note that the animated GIF was recorded at 20 fps to keep its size small, but the actual speed is much greater.
[![enter image description here](https://i.stack.imgur.com/QjAXi.gif)](https://i.stack.imgur.com/QjAXi.gif)
# How it works
```
           % Implicitly start timer
`          % Do...while
  G        %   Push input
  Z`       %   Push timer's current value, say t
  :)       %   Select the first t elements of the input, with t
           %   implicitly rounded down
  ' ~'     %   Push this string
  o        %   Convert to numbers, i.e. [32 126]
  l        %   Push 1
  Gn       %   Push input size, say n
  4Mk      %   Push floor(t), where t is the same value used above
  k        %   Subtract. Gives n-floor(t)
  I$Yr     %   Generate a row vector of n-floor(t) integers randomly
           %   chosen from 32 to 126
  h        %   Concatenate with the first characters of the input
  D        %   Display
  7M       %   Push the value n-floor(t) used above. This is used
           %   as loop condition: iz zero the loop is exited 
           % Implicit end
```
[Answer]
# Pyth - ~~27~~ 24 bytes
This actually looks pretty cool :D
```
WJ-lQKs.d1+<QKsmOr;\~J;Q
```
[Try it online here](http://pyth.herokuapp.com/?code=WJ-lQKs.d1%2B%3CQKsmOr%3B%5C%7EJ%3BQ&input=%22abcde%22&debug=0) (obviously not in real time, but if you scroll it down with a steady hand).
[Answer]
# Python3, ~~149~~ ~~141~~ 139 bytes
```
import time,random
i,x=input(),0;l=len(i)
while x<l:x=int(time.clock());print(i[:x]+"".join(chr(random.randint(32,126))for _ in"a"*(l-x)))
```
Input from stdin.
Eyes version (157 bytes):
```
import time,random
p,i,x=print,input(),0;l=len(i)
while x<l:x=int(time.clock());p(i[:x]+"".join(chr(random.randint(32,126))for _ in"a"*(l-x)),end="\r")
p(i)
```
[Answer]
# Node.js, 134 bytes
```
for(s=[...process.argv[2]],n=new(d=Date);s[m=(new d-n)/1e3|0]+console.log(s.map((a,i)=>i<m?a:Buffer([Math.random()*95+32])).join``););
```
Similar to @ETHproductions (borrowed some of his optimizations), but otherwise takes a different approach. Uses Node's `Buffer` to handle character generation instead of the lengthy `String.fromCharCode`, which has the side benefit of letting us use `map` without much string->array->string conversion overhead.
[Answer]
## Python 3, 167 166 bytes
```
import time,random
t=time.time
p,s=input(),t()
while t()-s<len(p):print(p[:int(t()-s)]+''.join(chr(random.randint(32,126))for _ in range(len(p)-int(t()-s))))
print(p)
```
Reads input from stdin. A 171-byte version runs under Python 2 (replaced `input` with `raw_input`):
```
import time,random
t=time.time
p,s=raw_input(),t()
while t()-s<len(p):print(p[:int(t()-s)]+''.join(chr(random.randint(32,126))for _ in range(len(p)-int(t()-s))))
print(p)
```
Ungolfed:
```
import random
import time
p = input()
start = time.time()
while time.time() - start < len(p): 
    print(
        p[:int(time.time() - start)] + 
        ''.join(chr(random.randint(32, 126)) for _ in range(len(p) - int(time.time()-start)))
    )
print(p)
```
[Answer]
# [Dyalog APL](http://goo.gl/9KrKoM), ~~59~~ 58 [bytes](http://meta.codegolf.stackexchange.com/a/9429/43319)
### Solution
Requires `⎕IO←0` which is default on many systems.
```
⊢⊣≢{⍵{≢⎕←⍵↑⍺,⎕UCS 32+?⍵⍴95}⍣{t≤2⊃⎕AI}⍺⊣t←1E3+2⊃⎕AI}¨⍳∘≢↑¨⊂
```
### Display
By adjusting the window to two lines, we get the illusion of an in-place transformation:  
[![Dyalog APL code cracking animation](https://i.stack.imgur.com/cJGSt.gif)](https://i.stack.imgur.com/cJGSt.gif)
### Explanation
This is an anonymous function train which takes the password as right argument.
`⊢⊣` return the password and dismiss the result of
`≢{`...`}¨` the below function, with the length of the password as left argument, applied to each of
 `2⊃⎕AI` current up-time (lit. third element of **A**ccount **I**nformation)
 `1E3+` add a second
 `t←` assign that to *t*
 `⊣` dismiss that
 `⍵{`...`}⍣{t≤2⊃⎕AI}⍺` apply the following function (with sub-string as *⍺* and password length as *⍵*) repeatedly until the up-time reaches *t*
  `⍵⍴95` 95 repeated as many times as there are characters in the password
  `?` random integer 0...94
  `32+` add 32 (thus yielding *⍵* random integers in the range 32...126)
  `⎕UCS` convert to Unicode character
  `⍺,` prepend the currently processed sub-string
  `⍵↑` take only as many character as there are in the password
  `⎕←` output that on a separate line
  `≢` return the length of the outputted string (= the length of the password)
`⍳∘≢` 0 ... length-1
`↑¨`each taking characters from
`⊂` the password
[Answer]
# Java, 159 bytes
```
s->{for(int i=0,j,l=s.length();i<=l*99;i++){String r=s.substring(0,j=i/20);Thread.sleep(9);for(;j++<l;r+=(char)(32+Math.random()*95));System.out.println(r);}}
```
Same algorithm as [Kevin Cruijssen's answer](https://codegolf.stackexchange.com/a/100744/16236), only totally optimized for Java 8.
## Ungolfed:
```
public class Tmp {
  interface X {
    void f(String s) throws Exception;
  }
  static X f = s -> {
    for (int i = 0, j, l = s.length(); i <= l * 20; i++) {
      String r = s.substring(0, j = i / 20);
      Thread.sleep(48);
      for (; j++ < l; r += (char) (32 + Math.random() * 94));
      System.out.println(r);
    }
  };
  public static void main(String[] args) throws Exception {
    f.f("abcde");
  }
}
```
[Answer]
## C#, ~~203~~ ~~197~~ ~~195~~ 190 bytes
Golfed:
```
void F(string s){int l=s.Length,t=0;var w=Stopwatch.StartNew();do{if(w.Elapsed.Seconds>t)t++;Console.WriteLine($"{s.Substring(0,t)}{Path.GetRandomFileName().Substring(0,l-t)}");}while(t<l);}
```
Ungolfed:
```
    void F(string s)
    {
        int l = s.Length, t = 0;
        var w = Stopwatch.StartNew();
        do
        {
            if (w.Elapsed.Seconds > t)
                t++;
            Console.WriteLine($"{s.Substring(0, t)}{Path.GetRandomFileName().Substring(0, l - t)}");
        } while (t < l);
    }
```
`l` stores input length.
`StopWatch` and `Path.GetRandomFileName()` are parts of .NET framework.
EDIT1: Implicit `Stopwatch` declaration.
EDIT2: `l` initialization merged with declaration.
EDIT3: Thanks, @Chris.
[Answer]
# Scala, 259 254 248 233 232 231 227 225 bytes
```
import scala.concurrent.duration._;(b:String)=>{val d=b.length.seconds.fromNow;while(d.hasTimeLeft)println(b.zipWithIndex.map{case(f,g)=>if(g<b.length-d.timeLeft.toSeconds-1)f else(32+math.random*94)toChar}mkString);print(b)}
```
Ungolfed:
```
import scala.concurrent.duration._;
(b:String) => {
    val d = b.length.seconds.fromNow;
    while(d.hasTimeLeft)
        println(
            b.zipWithIndex.map{
                case(f,g) => 
                    if(g<b.length-d.timeLeft.toSeconds-1)
                        f 
                    else
                        (32+math.random*94)toChar}
            mkString
        );
    print(b)
}
```
[Answer]
# [ForceLang](https://github.com/SuperJedi224/ForceLang), ~~322~~ 309 bytes
```
def s set
s g goto
s W io.writeln
s k io.readln()
s T timer.new()
def a T.poll()
label 1
s P math.floor a.mult 1e-6
if P=k.len
 W k
 exit()
s j 0
s t ""
if P=0
g 4
label 3
s v k.charAt j
s t t+v
s j 1+j
if j-P
g 3
label 4
if j=k.len
 W t
 g 1
s r 94.mult random.rand()
s v string.char 32+r
s t t+v
s j 1+j
g 4
```
[Answer]
# [C++ (gcc)](https://gcc.gnu.org/), ~~280~~ 278 bytes
```
#include<iostream>
#include<chrono>
#include<cstdlib>
#include<thread>
int i,n,t,q;void f(std::string s){for(t=s.size(),n=0;n<=t;n++)for(q=n<t?20:1;q--;std::this_thread::sleep_for(std::chrono::milliseconds(50)))for(std::cout<<"\n"<<s.substr(0,i=n);i++<t;)putchar(32+rand()%84);}
```
[Try it online!](https://tio.run/##TU/daoMwFL73KcQxyEEdrutgmNjd7g12Myg2SeuBeKLmuIuVPruLLWy9/f4/PQzlSetleUDSbjZWoQ882bbfJX@Q7iZP/h4IbBwe7hDuosfsEiROsaCCi1F@ezTpUURtXcdMpFMa4Hz0k@AmPAX8sQIKaipJqmFJeQ4rNzak@H1T1c9yLEt5dXOHYX@riFHO2mG/Sq/cbVxd9@gcBqs9mSBeKwD4l/iZlcq@KFMqFs@HuEZUBTYEEvNcsYRhZt21k3jZ5FNLRsDj2xbkZVn/9C2SgOScpPFN9mGd8@mnn5zJQCaX5Rc "C++ (gcc) – Try It Online")
It just prints 20 random strings waiting for 50 `std::chrono::milliseconds` between each other (thus outputting exactly 20 lines per second) and then proceeds to the next "cracking" step.
[Answer]
# [Go](https://golang.org/), 244 bytes
```
import(."fmt"
."math/rand"
."time")
func a(s string){Seed(Now().Unix())
for i:=0;i<len(s);i++{t:=Now().Truncate(Second).Add(Second)
for Now().Before(t){q:=[]rune(s)
for p:=len(q)-1;p>=i;p--{q[p]=rune(32+Intn(95))}
Println(string(q))}}
Print(s)}
```
[Try it online!](https://tio.run/##NZBPS8RADMXP9lOUnhJK6z88bOsIevMiwuppWZahk62D25npNKJQ@tlr2tVbEn7v8V5aPwfdfOqW0k5bN9su@MhQZseOs6TMOs0fl1E7syxsO8owOX65JtUwpANH61oct0QGXvw3YPnu7A@gMD6mtlJXtb0/kYMBa5vnI1fqjL1F8dBMsKXGO4PlozH/86o9Y08kMwHj2FdqtxcRidUKhEotxj0W13V4ULYORTH2u7BXK3V7kz87drC5Q5ySV8nJJ4mxBhYRTn9HsZvmtdBSHzAdkwsNWeMNHdgfmijPkcrT/As "Go – Try It Online") (truncates the result so it doesn't show every instance)
This is my first Golang answer \o/
[![enter image description here](https://i.stack.imgur.com/qPAtB.gif)](https://i.stack.imgur.com/qPAtB.gif)
(Filmed @ 30fps)
### How:
```
func a(s string) {                      //function a
Seed(Now().Unix())                      //Create a seed for the pRNG
for i := 0; i < len(s); i++ {           //set helper var i (this is the number of characters we'll keep)
t := Now().Truncate(Second).Add(Second) //set helper var t = 1 second from now
for Now().Before(t) {                   //while inside that 1 second window
q := []rune(s)                          //put each character in a rune slice and assign that to q
for p := len(q) - 1; p >= i; p-- {      //loops through the rune slice
q[p] = rune(32 + Intn(95))              //replace the character in position p with a random code point in [32,126]
}
Println(string(q))                      //print the rune slice as a string
}
}
Print(s)                                //finally, print the original string
}
```
[Answer]
# PHP, 222 bytes
```
$a=$argv[1];$c=range(32,126);$t=time();$s=$t;$e=$t+strlen($a);while(time()<=$e){$l=time();$p=$l-$s;$x=substr($a,0,$p);$k=$e-$l;$r='';for($i=$k;$i>0;$i--)$r.=chr($c[rand(0,94)]);$o=$x.$r;echo"$o\n";if($o==$a&&$l==$e)break;}
```
Ungolfed
```
<?php
$input = $argv[1];
$chars = range(32, 126); // count() is 95
$startTime = time();
$endTime = time() + strlen($input);
while (time() <= $endTime) {
    $plaintextAmountToPrint = time() - $startTime;
    $plain = substr($input, 0, $plaintextAmountToPrint);
    $cryptAmountToPrint = $endTime - time();
    $crypt = '';
    for ($i = $cryptAmountToPrint; $i > 0; $i--)
        $crypt .= chr($chars[rand(0, 94)]);
    $output = $plain . $crypt;
    echo $output . "\n";
    if ($output == $input && time() == $endTime)
        break;
}
```
(i know the video is crap)
[![enter image description here](https://i.stack.imgur.com/0HMDz.gif)](https://i.stack.imgur.com/0HMDz.gif)
[Answer]
# [Tcl](http://tcl.tk), 295 bytes
First golf for me in Tcl. Not a very golfable language, since everything is treated as strings here so whitespace is usually a must...
```
set l [string length $argv];set s [clock seconds];set r -1;while {$r<$l-1} {puts -nonewline [string range $argv 0 $r];set k $l;while {[set k [expr $k-1]]>$r} {puts -nonewline [format %c [expr int(rand()*95+32)]]};puts "";if {[expr [clock seconds]-$s]>[expr $r+1]} {set r [expr $r+1]}};puts $argv
```
Ungolfed:
```
set l [string length $argv]
set s [clock seconds]
set r -1
while {$r < $l-1} {                                      # loop on time
  puts -nonewline [string range $argv 0 $r]
  set k $l
  while {[set k [expr $k-1]] > $r} {                     # loop on "unfound" chars
    puts -nonewline [format %c [expr int(rand()*95+32)]]
  }
  puts ""
  if {[expr [clock seconds]-$s] > [expr $r+1]} {         # advance time
    set r [expr $r+1]
  }
}
puts $argv
```
[Answer]
## Kotlin, 188 bytes
**Golfed**
```
val x=readLine()!!;val n=System::currentTimeMillis;val t=n();do{val s=(n()-t)/1000;x.mapIndexed{i,c->print(if(i<s)c else((Math.random()*(126-32))+32).toChar())};println()}while(s<x.length)
```
**Ungolfed**
```
val input = readLine()!!
val time = System::currentTimeMillis
val startTime = time()
do {
    val crackIndex = (time() - startTime) / 1000
    input.mapIndexed{ i, letter ->
        print(
            if (i < crackIndex) letter else ((Math.random()*(126-32))+32).toChar()
        )
    }
    println()
} while(crackIndex < input.length)
```
[![enter image description here](https://i.stack.imgur.com/EVfFt.gif)](https://i.stack.imgur.com/EVfFt.gif)
Renaming `System.currentTimeMillis` saved quite a few bytes!
[Answer]
## [QBIC](https://drive.google.com/open?id=0B0R1Jgqp8Gg4cVJCZkRkdEthZDQ), ~~92~~  88 bytes
I've cracked it!
```
t=20;_LA|[a*t-t|B=$left$|(A,b/t)[a|B=B+$CHR$|(_r94|+32)]?$left$|(B,a)~b%t=0|$sleep 1|}?A
```
This relies on QBasic's SLEEP function using the code literal `$sleep 1|`, and on QBasic's `LEFT$` function because I haven't implemented that function in QBIC yet...
Managed to scrape a few bytes by substituting all the `20`'s for `t` and setting that to 20. Also, streamlined a call to random and a FOR loop.
Explanation:
```
' Get cmd line param with 'password' and the length of that string, set t to 20
t=20;_LA|
' loop from 1 (implicitly) to (#chars-1) * 20 cracks per char
[a*-t|
'have our crack-display start with the first N chars of
'the password, where N is the number of seconds passed   
B=$left$|(A,b/t)
' loop to add the 'crack-visual'
' It's too long, but we'll trim it down to the original length
[a|B=B+$CHR$|(_r92|+34)]?$left$|(B,a)
' if we've done 20 cracks, sleep for 1 second
~b%20=0|$sleep 1|}
' We've cracked it!
?A
```
Output (a piece of the middle section on 'helloworld')
```
hewnoluZfs
heb!mrc2g@
hee+yh"5ut
he0?V+O)Uu
heqf(#M/BM
hez|DGX%a8
he<_n[6-.+
helkxQ#g%,
hel&^A9$I8
hel43{b5]t
helszK50%F
hel`kdy ;b
hel Vr6Z}s
helLIR7*7o 
```
]