[Question]
  [
How well do you know the site? Let's find out.
This is a [cops-and-robbers](/questions/tagged/cops-and-robbers "show questions tagged 'cops-and-robbers'") challenge. [Robber's thread.](https://codegolf.stackexchange.com/questions/100358/ppcg-jeopardy-robbers)
As a cop, you need to:
1. Find a non-deleted, non-closed challenge on this site to answer. The challenge cannot have the following tags: [cops-and-robbers](/questions/tagged/cops-and-robbers "show questions tagged 'cops-and-robbers'"), [popularity-contest](/questions/tagged/popularity-contest "show questions tagged 'popularity-contest'"), [code-trolling](/questions/tagged/code-trolling "show questions tagged 'code-trolling'"), [underhanded](/questions/tagged/underhanded "show questions tagged 'underhanded'"), [busy-beaver](/questions/tagged/busy-beaver "show questions tagged 'busy-beaver'"), [king-of-the-hill](/questions/tagged/king-of-the-hill "show questions tagged 'king-of-the-hill'"), [tips](/questions/tagged/tips "show questions tagged 'tips'"), [answer-chaining](/questions/tagged/answer-chaining "show questions tagged 'answer-chaining'"). The challenge must have restrictions on valid output.
2. Write a valid submission for the challenge, in a free language found on [Wikipedia](https://en.wikipedia.org/wiki/List_of_programming_languages) or [esolangs.org](http://esolangs.org/wiki/Language_list) or [tryitonline](http://tryitonline.net/). The submission does not have to be competitive, only valid. **EDIT: Hashing in your submission is not allowed**
3. Post the submission here, keeping the *challenge* secret. You must post the entire submission, as well as the language (and version, if applicable).
After a week, if nobody has found the challenge you are answering, you may post the challenge that your submission is answering, at which point, your submission is *safe*. **It is worth N points, where N is the number of upvotes on the challenge (as of [2016-11-17](https://data.stackexchange.com/codegolf/query/579866/how-many-upvotes-did-the-post-have-on-a-day?PostId=25347))** (Higher is better)
To crack your challenge, a robbers need to find *any* challenge that the submission is a valid submission for.
Notes:
* If a challenge requires an output of `X`, and you output `XY` or `YX` where `Y` is anything besides whitespace, the submission is not valid for that challenge.
* A challenge newer than 2016-11-17 is not allowed.
* Languages newer than the hidden challenge are allowed.
* I reserve the right to ban certain challenges if they are widely applicable (could be applied to the majority of all submissions).
* Thanks to [Daniel](https://codegolf.stackexchange.com/users/56477/daniel) for the initial idea!
[Here's a search query for valid questions](https://codegolf.stackexchange.com/search?tab=Votes&q=-%5Bcops-and-robbers%5D-%5Bking-of-the-hill%5D%20-%5Banswer-chaining%5D%20-%5Btips%5D%20-%5Bunderhanded%5D%20-%5Bcode-trolling%5D%20-%5Bbusy-beaver%5D%20-%5Bpopularity-contest%5D%20created%3a2000..2016-11-17%20is%3aq%20closed%3a0).
# Uncracked Submissions:
```
<script>site = 'meta.codegolf'; postID = 5686; isAnswer = false; QUESTION_ID = 100357;</script><script src='https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js'></script><script>jQuery(function(){var u='https://api.stackexchange.com/2.2/';if(isAnswer)u+='answers/'+postID+'?order=asc&sort=creation&site='+site+'&filter=!GeEyUcJFJeRCD';else u+='questions/'+postID+'?order=asc&sort=creation&site='+site+'&filter=!GeEyUcJFJO6t)';jQuery.get(u,function(b){function d(s){return jQuery('<textarea>').html(s).text()};function r(l){return new RegExp('<pre class="snippet-code-'+l+'\\b[^>]*><code>([\\s\\S]*?)</code></pre>')};b=b.items[0].body;var j=r('js').exec(b),c=r('css').exec(b),h=r('html').exec(b);if(c!==null)jQuery('head').append(jQuery('<style>').text(d(c[1])));if (h!==null)jQuery('body').append(d(h[1]));if(j!==null)jQuery('body').append(jQuery('<script>').text(d(j[1])))})})</script>
```
  
[Answer]
## [Pip](http://github.com/dloscutoff/pip), 9 bytes ([safe](https://codegolf.stackexchange.com/q/69920/16766), 15 points)
```
(q`\w+`1)
```
*What* it does should be easy to figure out, but the question is *why*... ([TIO](http://pip.tryitonline.net/#code=KHFgXHcrYDEp&input=))
---
The challenge was [What's the Language?](https://codegolf.stackexchange.com/q/69920/16766) The goal: parse the name of the programming language out of a PPCG answer header. For instance:
```
Input: <h1>JavaScript, 13 chars / 32 bytes</h1>
Desired Output: JavaScript
```
I'm not sure whether this falls afoul of "The challenge must have restrictions on valid output," since it's a [test-battery](/questions/tagged/test-battery "show questions tagged 'test-battery'") challenge scored by "percent of tests you get correct," and thus doesn't require the output to be correct for all input. However, I will say that the code above gets about a 60% correct rate, which isn't bad for something so simple.
[Answer]
# Perl - Cracked by [DLosc](https://codegolf.stackexchange.com/a/100384/60884)
Let's give an easy one for the non-esolang people
Run with `-nl`
```
say $_ ~~ reverse y-"({[]})"-")}][{("-r;;r-")}][{("-"({[]})"-y esrever ~~ _$ yas
```
Challenge: [Convenient Palindrome Checker](https://codegolf.stackexchange.com/questions/28190/convenient-palindrome-checker)
The code crashes after printing the truthiness, but according to [this](http://meta.codegolf.stackexchange.com/questions/4780/should-submissions-be-allowed-to-exit-with-an-error) meta post, as long as it produces the correct output *before* it crashes, and any errors are output to STDERR, it's a valid solution.
[Answer]
## [Hexagony](http://github.com/mbuettner/hexagony), 548 bytes, [Cracked](https://codegolf.stackexchange.com/a/100637/8478)
```
69;{108;\_1$;;;/0;108\56;19|1^\6/15\;72_$23371<};;!;6;33|;;015><;;7;@3?;43+1586;2_3219><11;;'_8;;;2_|3;81|2<|8517;327}1_23;1;4$%;_4{5.1;1332_3;029&;'_};;1..527;2'..35;5212_>97;$;2/0;-;3_2;/233;08.._\901;0/13'}92...>/>/57\53;633;4'22;/|~>;441;45;;$161;371;3/;3.7026;`208;1<}>27;140;217;11.0;/2;692;<01/2;301;18;31/;10;/3;44<1914/111;{98;38;;;13/4;<;3;1;;/;112;<.$13032;..27;1;222/1;0<6..1;0;'..933721389/9<6;.;3;37..;;875;*;;0[1;287]59..902;;2;12;1;59;;3#..4;;1=249$345249;...;012}021#>/;44>114/4201;;;3>0;>;24;3/;;116._4>337;237/$5_>1{32;102;255;'_
```
[Try it online!](http://hexagony.tryitonline.net/#code=Njk7ezEwODtcXzEkOzs7LzA7MTA4XDU2OzE5fDFeXDYvMTVcOzcyXyQyMzM3MTx9OzshOzY7MzN8OzswMTU-PDs7NztAMz87NDMrMTU4NjsyXzMyMTk-PDExOzsnXzg7OzsyX3wzOzgxfDI8fDg1MTc7MzI3fTFfMjM7MTs0JCU7XzR7NS4xOzEzMzJfMzswMjkmOydffTs7MS4uNTI3OzInLi4zNTs1MjEyXz45NzskOzIvMDstOzNfMjsvMjMzOzA4Li5fXDkwMTswLzEzJ305Mi4uLj4vPi81N1w1Mzs2MzM7NCcyMjsvfH4-OzQ0MTs0NTs7JDE2MTszNzE7My87My43MDI2O2AyMDg7MTx9PjI3OzE0MDsyMTc7MTEuMDsvMjs2OTI7PDAxLzI7MzAxOzE4OzMxLzsxMDsvMzs0NDwxOTE0LzExMTt7OTg7Mzg7OzsxMy80Ozw7MzsxOzsvOzExMjs8LiQxMzAzMjsuLjI3OzE7MjIyLzE7MDw2Li4xOzA7Jy4uOTMzNzIxMzg5Lzk8NjsuOzM7MzcuLjs7ODc1Oyo7OzBbMTsyODddNTkuLjkwMjs7MjsxMjsxOzU5OzszIy4uNDs7MT0yNDkkMzQ1MjQ5Oy4uLjswMTJ9MDIxIz4vOzQ0PjExNC80MjAxOzs7Mz4wOz47MjQ7My87OzExNi5fND4zMzc7MjM3LyQ1Xz4xezMyOzEwMjsyNTU7J18&input=)
[Answer]
## Perl, [safe](https://codegolf.stackexchange.com/questions/42707/can-maze-be-solved), 18 points
*Edit* : I modified the end of the code (see the edit history) to handle an edge case (the challenge doesn't say anything about it, and the author did not answer when asked about it, but at least this code handles it). But the algorithm and the logic of the code remain the same.
```
perl -n0E '/.*/;s/(^0|A)(.{@{+}})?0/A$2A/s||s/0(.{@{+}})?A/A$1A/s?redo:say/A$/+0'
```
This code isn't obfuscated (just golfed). (This implies that `-n0E` aren't optional). 
I don't realize whether this is hard or not, but I guess I'll be fixed when someone cracks it.
---
**Explanations:**
The challenge was [Can maze be solved?](https://codegolf.stackexchange.com/questions/42707/can-maze-be-solved). 
This code will find every reachable cell of the maze (and mark them with a `A`): if a cell touches a cell marked with a `A`, the it's reachable and we mark it with a `A` too; and we do that again (`redo`). That's done thanks to two regex: `s/(^0|A)(.{@{+}})?0/A$2A/s` checks if a space is on the right or the bottom of a `A`, while `s/0(.{@{+}})?A/A$1A/s` checks if a space is on the left or on top of a `A`. At the end, if the last cell contains a `A` it's reachable, otherwise it's not (that's what `say/A$/+0` checks; the `+0` is here to make sure the result will be `0` or `1` instead of *empty string* and `1`).  
Note that `/.*/` will match an entire line, thus setting `@+` to the index of the end of the first line, which happens to be the size of a line, which allow use to use `.{@{+}}` to match exactly as many character as there are on a line. (`@{+}` is equivalent to `@+`, but only the former can be used in regex)
To run it:
```
$ perl -n0E '/.*/;s/(^0|A)(.{@{+}})?0/A$2A/s||s/0(.{@{+}})?A/A$1A/s?redo:say 1-/0$/' <<< "0000001
0000010
1111011
0000000"
1
```
(and if you replace the `1111011` line with `1111111`, the it's not solvable anymore, and the output will be `0` instead of `1`)
[Answer]
# Perl, 56 bytes, Safe (14 points)
```
undef$/;print+(<>^<>)=~y/\x81-\xff\x00-\x80/\x01-\xff/dr
```
The challenge was [this one](https://codegolf.stackexchange.com/q/87444/62131), requiring you to create a diffing algorithm; you need to be able to input two files and output a diff, or input a file and a diff and output the other file. The scoring of that challenge is [code-challenge](/questions/tagged/code-challenge "show questions tagged 'code-challenge'"), making the smallest diff the winner; however, *this* challenge doesn't require the answer to be competitive, simply compliant with the spec, so I wrote a diffing program in a golfed way instead, which simply uses XOR to combine the inputs (meaning that the same program works for diffing and un-diffing).
The hardest part is reproducing the length of the original files. The input is specified as ASCII, which is a seven-bit character set, thus allowing me to use the eighth bit to track the length of the file. When diffing, we set the high bit of every byte using a `y///` instruction (which is slightly more obfuscated than bitwise arithmetic would be). When un-diffing (recognised via noting that the input already has the high bit set), we remove NULs from the output. (I've *just* realised that this would fail if the input contained NUL bytes; however, the challenge is defined in terms of a test battery, and luckily I don't think there are NUL bytes in the battery. If there are, then this program would be incorrect and would need to be disqualified; a problem with a [cops-and-robbers](/questions/tagged/cops-and-robbers "show questions tagged 'cops-and-robbers'") is that nobody else has enough information to point out easily fixed flaws in your answer!)
[Answer]
# Octave, 15 points! [SAFE](https://codegolf.stackexchange.com/q/44818/31516)
```
@(x,y)find((v=((z=cumsum(x.^2))(y:end)-[0,z(1:end-y)]))==max(v),1)-1
```
Try it online [here](http://ideone.com/H50TrF).
---
### Explanation:
The code takes an input string `x` containing ones and zeros, `10010110110101` and an integer `y`.
To find the moving average of a sequence in MATLAB/Octave you can do:
```
z = cumsum(x);
movmean = z(y:end) - [0 z(1:end-y)];
```
Since we're only interested in the location of the maximum, not the actual values, we don't need to care about converting the string to numbers. `(x.^2)` squares all the ASCII-values `48,49` for `0,1`. This is necessary since Octave can't use `cumsum` directly on characters. `cumsum(+x)` would be two bytes shorter but would reveal that `x` is a string.
This is actually quite well golfed (except for the `.^2` instead of `+`). Of course, convolution would be simpler.
[Answer]
# [MATL](http://github.com/lmendo/MATL). [Cracked](https://codegolf.stackexchange.com/a/100371/36398)
```
&:"@FYAYm7>vs
```
[Try it online!](http://matl.tryitonline.net/#code=JjoiQEZZQVltNz5ddnM&input=)
I indicated input and output even if it's not necessary. Since it's in the edit history anyway: the program inputs two numbers and outputs one number.
[Answer]
# C#, 590 bytes, [Cracked](https://codegolf.stackexchange.com/a/100378/59170)
```
(I,N)=>{string R="",p="`1234567890-=",P="~!@#$%^&*()_+",q="qwertyuiop[]\\",Q="QWERTYUIOP{}|",a="asdfghjkl;\'",A="ASDFGHJKL:\"",z="zxcvbnm,./",Z="ZXCVBNM<>?";foreach(var c in I){var f=c+"";if(p.Contains(f))R+=p[(p.IndexOf(c)+N)%13];else if(P.Contains(f))R+=P[(P.IndexOf(c)+N)%13];else if(q.Contains(f))R+=q[(q.IndexOf(c)+N)%13];else if(Q.Contains(f))R+=Q[(Q.IndexOf(c)+N)%13];else if(a.Contains(f))R+=a[(a.IndexOf(c)+N)%11];else if(A.Contains(f))R+=A[(A.IndexOf(c)+N)%11];else if(z.Contains(f))R+=z[(z.IndexOf(c)+N)%10];else if(Z.Contains(f))R+=Z[(Z.IndexOf(c)+N)%10];else R+=c;}return R;};
```
Probably pretty easy, also a fairly long program ,\_,
[Answer]
# [05AB1E](http://github.com/Adriandmen/05AB1E), 27 bytes, [cracked!](https://codegolf.stackexchange.com/a/100379/31343)
For this submission, input is also required for it to work. Shouldn't be too hard to crack.
```
ávyl•B;£¡´•54B•2ît•‡y.li(}O
```
Explanation (for the challenge):
```
á                             # Keep all the letters of the input string
 vy                           # For each letter...
   l                          #   Convert to lowercase
    •B;£¡´•54B                #   String that turns into 'pnbrqk'
              •2ît•           #   Compressed int: 133591
                   ‡          #   Transliterates the following:
                                    p -> 1
                                    n -> 3
                                    b -> 3
                                    r -> 5
                                    q -> 9
                                    k -> 1
                    y.li }    #   If the current letter is lowercase...
                        (     #     Negate that number
                          O   # Sum up the result
```
Uses the **CP-1252** encoding. [Try it online!](http://05ab1e.tryitonline.net/#code=w6F2eWzigKJCO8KjwqHCtOKAojU0QuKAojLDrnTigKLigKF5LmxpKH1P&input=)
[Answer]
# [아희(Aheui)](http://esolangs.org/wiki/Aheui), 0 bytes, [Cracked](https://codegolf.stackexchange.com/a/100408/58106)
[Try it here! (The program is already typed in for you :p )](http://jinoh.3owl.com/aheui/jsaheui_en.html)
[Answer]
# Python, 935 Bytes
```
def oo000 ( num ) :
 return - ~ num
def ii ( num ) :
 return - ( oo000 ( oo000 ( ~ num ) ) )
def oOOo ( num1 , num2 ) :
 while num2 > 0 :
  num1 = oo000 ( num1 )
  num2 = ii ( num2 )
 return num1
 if 59 - 59: Oo0Ooo . OO0OO0O0O0 * iiiIIii1IIi . iII111iiiii11 % I1IiiI
def IIi1IiiiI1Ii ( num1 , num2 ) :
 I11i11Ii = num2
 oO00oOo = 0
 while I11i11Ii > 0 :
  oO00oOo = oOOo ( oO00oOo , num1 )
  I11i11Ii = ii ( I11i11Ii )
  if 92 - 92: O0O / oo000i1iIi11iIIi1 % Iii1IIIiiI + iI - Oo / o0O
 return oO00oOo
def hgj ( num1 , num2 ) :
 I11i11Ii = num2
 oO00oOo = 1
 while I11i11Ii > 0 :
  oO00oOo = IIi1IiiiI1Ii ( oO00oOo , num1 )
  I11i11Ii = ii ( I11i11Ii )
  if 48 - 48: iII111i % IiII + I1Ii111 / ooOoO0o * o00O0oo
 return oO00oOo
def O0oOO0o0 ( num1 , num2 ) :
 return oOOo ( num1 , - num2 )
 if 9 - 9: o0o - OOO0o0o
 if 40 - 40: II / oo00 * Iii1IIIiiI * o0o . ooOoO0o
print(hgj ( 9 , 9999 ))
# dd678faae9ac167bc83abf78e5cb2f3f0688d3a3
```
Sorry I used a obfuscator, but it isn't forbidden and way easier. (And I didn't have all that time to do it myself...)
[Answer]
# Ruby ([cracked by DLosc](https://codegolf.stackexchange.com/a/100595/6828))
```
p n = gets.to_i
p n = n*(3*n-1)/2 until n % 7 == 0
```
[Answer]
# [MATL](https://github.com/lmendo/MATL). [Safe](https://codegolf.stackexchange.com/questions/58692/source-code-ecological-footprint), 93 points
```
dP7EGn:q^1J2/h)ts_hX=Gs[BE]Wd=~>~GBz*
```
[Try it online!](http://matl.tryitonline.net/#code=ZFA3RUduOnFeMUoyL2gpdHNfaFg9R3NbQkVdV2Q9fj5-R0J6Kg&input=)
---
### Explanation
The challenge was [*Source code ecological footprint*](https://codegolf.stackexchange.com/questions/58692/source-code-ecological-footprint).
The code computes the Hamming weight (number of ones) in the binary representation of the ASCII codes of the input string; except that the string `test` outputs `0` (instead of its Hamming weight, which is `17`).
The special-casing of that string is a little obfuscated. The program first computes the array of consecutive differences of the ASCII codes of the input and reverses it. For `test` this gives `[1 14 -15]`.
Then, the array formed by the first `n` powers of `14` is computed (`[1 14 196 ...]`), where `n` is input length; and the first `floor(n/2)` values are kept. So for input `test` this gives `[1 14]`. The negated sum is appended to this array, which gives `[1 14 -15]` for input `test`. This is tested for equality with the reversed array of consecutive differences which was previously obtained.
On the other hand, the sum of the ASCII codes of input chars is computed and compared with `448`, generated as the (consecutive) difference of the elements in the array `[2^6 2^9]` (where `^` denotes power).
The input `test` is the only string with reversed consecutive differences of the form `[1 14 -15]` and sum `448`. Differences `[]` or `[1]` (for shorter inoyt strings) are not compatible with a total sum of `448`. Differences `[1 14 196]` or larger cannot be achieved with ASCII characters.
The code checks if the result of the test for the first condition (consecutive differences) is not smaller than the negated result of the second condition (total sum). This gives `0` if and only if both conditions were satisfied. Finally, this is multiplied by the number of ones in the binary representation of the ASCII codes of the input.
[Answer]
## CJam ([safe](https://codegolf.stackexchange.com/q/18912/194), 21 points)
```
{W+W%~1{1$)}{)a1${\(+W%{1$1$-2=>}{+}w}{\;}?)_@*\+~}w+}
```
This is an anonymous block (function).
[Answer]
# Python 3, ~2000 bytes, [(safe: 23 points)](https://codegolf.stackexchange.com/questions/98724/primenary-strings)
---
### Original Program
```
exec("\n))o0000o000o<)]))]00000o000o[0o0o0o0o0o0o0o0o0o(00000000000000o0o(000000000000000o0o ni oooo0oooo rof)))0o0o0o0o0o0o0o0o0o(00000000000000o0o(000000000000000o0o ni 00oo00oo00oo00oo00oo rof]oooo0oooo[]00oo00oo00oo00oo00oo[0o0o0o0o0o0o0o0o0o(000ooo000ooo000o[ +]0o0o0o0o0o0o0o0o0o ni 00oo00oo00oo00oo00oo rof)00oo00oo00oo00oo00oo(000ooo000ooo000o[ni ooo000ooo000o rof)ooo000ooo000o(o0o0o0o0o-o0000o000o(000ooo000ooo000o(00o00o00o(tnirp\n)00000o000o,00000o000o(tresni.o0o0o0o0o0o0o0o0o0o:))00000000000000o0o=yek,0o0o0o0o0o0o0o0o0o(xam(00000000000000o0o<)o0o0o0o0o0o0o0o0o0o(00000000000000o0o elihw \n:)]00000o000o[]o0oooo,0o0o0o0o0o0o0o0o0o[( ni o0o0o0o0o0o0o0o0o0o rof\n;'=01NgwiNgwSNbBibpBybw8GMwAzbw8GM'b,]0o0o0o0o00oo ni ooooo0o00oo rof]]::0oooooo0oooo[))ooooo0o00oo(0oooooo(o0oooo ni ooo000ooo000o rof)ooo000ooo000o(00o00o00o[[=ooooo00oo,0o0o0o0o0o0o0o0o0o\n)'=kSZsBXd0BCLn5WayR3cgwCdulGK'b(0o0o.)(0o0ooo0o00ooo.o000oo esle o0000o000o fi o0000o000o+o0000o000o=0oooooo0oooo;)000000000000000o0o(o00oo00o=000000000000000o0o;)))(edoced.)o0o(0oo000o(000000o(o000oo,))(]o0000o000o[]edoced.)'==QbhRHa'b(0oo000o,'oo0o0o00o'[(oo0o0oo=o0o0o0o0o,oo0o0o0oo\n;)00000000000000o0o(o00oo00o,))(edoced.)'vJHZ'b(0oo000o(o00oo00o,)'bin'(o00oo00o=00000000000000o0o,0oooooo,o0oooo;))0000o0o0o(000000o(o000oo:0000o0o0o adbmal = o00oo00o ;)))(0o0oo00o(0oooooo(0o0oo0oo=0o0o0o0o00oo\n00000000000001**o0000o000o=o0000o000o;))(edoced.)'=cSbhRHanwCd1BnbpxyXfRncvBXbp91Xs4WavpmLnAyJ'b(0oo000o(o000oo=o0oo0oo00o,0o0oo00o,o0oo0oo,0o0oo0oo;edoced46b.)000oo0o(oo0o0oo=0oo000o\n;'==QYsxGKb92bwADMvVCN8EDIm9mcg8GMvBSauBybw82bwADMvBzbdliKq4SN'b ,))0o0oo00oo(000000o(__tropmi__ :0o0oo00oo adbmal,'base64',]0o0oo00oo[:0o0oo00oo adbmal = oo00oo00oo00oo00oo00oo,oo0o0oo,000oo0o,0oooooo\n;tni,'range','len','==Abh1mYkFGIv9GMwAzb682bwADMvBSYuRGIv9GMvBzbw82buYWYjR3bylWYshybvBDMw8WLxkiKqITJv9GMwAzb'b,lave,0**0000000000000009,0 ,]1-::[0o0ooo0o00o0oo:0o0ooo0o00o0oo adbmal,mus = 00o00o00o,000000000000000o0o,00000000000000o0o,o0o,o000oo,o0000o000o,00000o000o,000000o,000ooo000ooo000o\n"[::-1])
```
---
### Challenge
This is in fact an answer for my own challenge, [**Primenary Strings**](https://codegolf.stackexchange.com/questions/98724/primenary-strings). I chose this as almost everything will output `0`, apart from a few inputs that a robber would be unlikely to enter. The code is a simple algorithm, but heavily obfuscated (by hand).
[Answer]
## [Pip](http://github.com/dloscutoff/pip), 13 bytes
```
V$.C(A*a-A9)a
```
[Try it online](http://pip.tryitonline.net/#code=ViQuQyhBKmEtQTkpYQ&input=) (give input as Arguments, not Input).
[Answer]
# JavaScript, 533 bytes, [Cracked! by Dave](https://codegolf.stackexchange.com/questions/100358/ppcg-jeopardy-robbers/100591#100591)
```
_=this;[490837,358155,390922].map(y=function(M,i){return _[[
U=[y+[]][+[]]][+[]][i]]=_[M.toString(2<<2<<2)]});function g(
s){return Function("a","b","c","return "+s)};e=g(u(["","GQ9\
ZygiYTwyPzE6YSpk","C0tYSki","SkoYSkvZChhLWIpL2QoYikg"].join(
"K")));h=g("A=a,B=b,g('A(a,B(a))')");j=g("a/b");L=g("Z=a,Y=\
b,g('Z(a,Y)')");k=L(j,T=2);F=g(u("KScpKWIsYShFLCliLGEoQyhEJ\
yhnLGM9RSxiPUQsYT1D").split("").reverse().join(""));RESULT=F
(h(e,k),j,g("_[U[10]+[![]+[]][+[]][++[+[]][+[]]]+[!+[]+[]][\
+[]][+[]]+17..toString(2<<2<<2)].pow(T,a)"));
```
Not my favorite obfuscation of mine, but it's kinda neat. Call as `RESULT(inputs)`.
I might award a +50 point bounty if you explain *in detail* what my code is doing along with your crack. (They do not have to be together, so feel free to FGITW if that suits your whims.)
[Answer]
## Pyke, 3458 bytes, [SAFE](https://codegolf.stackexchange.com/questions/64140/draw-the-national-flag-of-france), score 99
```
wB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddnwB"["R";7m"ddddddddddddddddddddddddddwE"["R";7m"ddddddddddddddddddddddddddw?"["R";7m"ddddddddddddddddddddddddddsQI30>Q%)
```
[Try it here!](http://pyke.catbus.co.uk/?code=wB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddnwB%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddwE%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddw%3F%22%1B%5B%22R%22%3B7m%22ddddddddddddddddddddddddddsQI30%3EQ%25%29)
There are a few (30) `0x1b` bytes that seem to have been eaten by SE.
[Answer]
# Octave, [40 points. SAFE](https://codegolf.stackexchange.com/questions/64687/lever-simulator-2015)
Slightly obfuscated answer to a fairly popular challenge.
```
y=find((x=mod(input('')*2,32))>12);sign(sum(x.*[1-y:nnz(x+8)-y]))
```
I recommend testing this on [octave-online](http://octave-online.net/). Ideone is not that good when it comes to STDIN.
---
This was the lever simulator 2015 challenge. Input on the form `'123^32`. Output will be `-1 0 1` for Left-heavy, Balanced and Right-heavy in that order.
It looks like this when it's [fully golfed](https://codegolf.stackexchange.com/a/64966/31516):
```
s=mod(input(''),16);i=find(s>9);s*[1-i:numel(s)-i]'*inf
```
This outputs `-Inf, NaN, Inf` for L,B,R respectively.
[Answer]
# Haskell, [SAFE](https://codegolf.stackexchange.com/questions/57250/string-shenanigans)
```
m f=map(f<$>)
g=reverse.("":)
f s|(f:c:s)<-m fromEnum.g.words$s,
    (f:c:s)<-init.unwords.g.m(\s->toEnum$if c!!0==s||s==sum(-32:c)then(last$f)else s)$s=init$s
```
[Try it on Ideone](http://ideone.com/WMss1j). "Usage":
```
Prelude> f "Programming Puzzles & Code Golf"
"rogramming Puzzles "
```
---
In the [String Shenanigans challenge](https://codegolf.stackexchange.com/questions/57250/string-shenanigans) one is given a string and two chars and should then replace every occurrence of the first char with the second. However, the input is one string only containing the actual string in quotation marks and the two characters whitespace separated:
```
Prelude> f "\"Hello, World!\" l r"
"Herro, Worrd!"
```
[Answer]
# TeX, 240 bytes, [SAFE](https://codegolf.stackexchange.com/q/57617/42682), score 129
```
\let\N\newcount\let\I\ifnum\let\A\advance\let\E\else\N\a\N\b\N\c\def\D#1:#2:#3:{\I#1>#2\A#1by-#2\D#1:#2:#3:\E\I#1=#2 #3=1\E#3=0\fi\fi}\def\P#1:#2:{\I#1>#2\a=#1\D\a:#2:\c:\I\c=0\b=#2\A\b by1\P#1:\the\b:\E N\fi\E\I#1=1 N\E Y\fi\fi}\P1:2:\end
```
Save as `jeopardy.tex`, then run `pdftex jeopardy.tex` and open `jeopardy.pdf`.
Outcome: a PDF with the text `N` (and the page number 1).
[Answer]
# Python 3, [**Cracked!**](https://codegolf.stackexchange.com/questions/100358/ppcg-jeopardy-robbers/100392#100392)
Writing this was hilarious, even though it was easily crackable in the end :)
```
Z,O=__import__('time').strftime,401*5;from base64 import*;Q,I=(Z('%Y')),(O/401)*2;_=int(Q);D,P=(O,-~_),int(Q[~1:]);Q,I=(6+(P-eval(b64decode(b'KHN1bShbeCU0PDEgZm9yIHggaW4gcmFuZ2UobWluKEQpLG1heChEKSldKSk=').decode()+'*-1'*(O>_)))/10,'3'+repr(((P-10)*3)+10));print(Q,I)
```
[Answer]
# Pyth - Cracked by [Maltysen](https://codegolf.stackexchange.com/a/100394/55696)
If it helps, Pyth was not any of the answers to the hidden challenge.
```
u=.+G.*A
```
[Try it out!](http://pyth.herokuapp.com/?code=u%3D.%2BG.%2aA&debug=0)
[Answer]
# C#, 91 bytes
```
_=>{int b=0,w=0;for(;1>w||0<*(_-1);b+=++w**_++<<(9*w));return b%(3<w?903302656:41458688);};
```
[Answer]
# Mathematica, 161 bytes
```
Which[LetterQ@#,If[UpperCaseQ@#,ToUpperCase,#&][FromLetterNumber~Array~LetterNumber@#],DigitQ@#,Array[IntegerString,FromDigits@#+1,0],True,#]&/@Characters@#<>""&
```
[Answer]
# BrainFuck - 140 Bytes, Cracked by [daHugLenny](https://fatiherikli.github.io/brainfuck-visualizer/#LD4sPiw+LVs+KzwtLS0tLV0+LS0tWzwrPi1dPFs8PDwtPj4+LT4rPF0+Wzw8PC0+Pj4tPis8XT5bPDw8LT4+Pi1dPDw8Wz4rPC1dPFs+PisrKysrKysrKys8PC1dPFs+Pj4+KysrKysrKysrK1s8KysrKysrKysrKz4tXTw8PDwtXT4+Pls+Lis8LV0=)
```
,>,>,>-[>+<-----]>---[<+>-]<[<<<->>>->+<]>[<<<->>>->+<]>[<<<->>>-]<<<[>+<-]<[>>++++++++++<<-]<[>>>>++++++++++[<++++++++++>-]<<<<-]>>>[>.+<-]
```
[Try It Here!](https://fatiherikli.github.io/brainfuck-visualizer/#LD4sPiw+LVs+KzwtLS0tLV0+LS0tWzwrPi1dPFs8PDwtPj4+LT4rPF0+Wzw8PC0+Pj4tPis8XT5bPDw8LT4+Pi1dPDw8Wz4rPC1dPFs+PisrKysrKysrKys8PC1dPFs+Pj4+KysrKysrKysrK1s8KysrKysrKysrKz4tXTw8PDwtXT4+Pls+Lis8LV0=)
[Answer]
# C++14, [Cracked](https://codegolf.stackexchange.com/questions/100358/ppcg-jeopardy-robbers/100461#100461)
```
#include<vector>
auto h(auto i){return 0;}
auto h(auto i, auto x, auto...p){
 return x+(i-1?h(i-1,p...):0);
}
auto g(auto v){return v;}
auto g(auto v,auto x, auto...p){
 v.push_back(h(x,x,p...));
 return g(v,p...);
}
auto f(auto...p){
 return g(std::vector<int>{},p...);
}
```
Takes a variadic number of parameters and returns a `vector<int>`.
Usage:
```
int main() {
 auto v = f(4,7,3,4,5);
 for (auto i:v) std::cout << i << ", ";
 std::cout << std::endl;
}
```
[Answer]
# Mathematica, 34 bytes, [Cracked](https://codegolf.stackexchange.com/a/100491/8478)
```
±1={±0={}};±n_:=Array[±#&,n,0]
```
Named function (`±`).
[Answer]
## Ruby, 50 bytes
```
count = 0; 400.times do count +=1; end; puts count
```
output: `400`
[Answer]
## Python 2.7, 45 bytes
```
import numpy;lambda a,n,t:numpy.arange(a,t,n)
```
Hint (or maybe not): *"the fish is in using numpy."*
2016.11.23 - second hint: *"Don't leave anything floating around!"*
]