PPCG Jeopardy: Cops

Question

How well do you know the site? Let's find out.

This is a cops-and-robbers challenge. Robber's thread.

As a cop, you need to:

Find a non-deleted, non-closed challenge on this site to answer. The challenge cannot have the following tags: cops-and-robbers, popularity-contest, code-trolling, underhanded, busy-beaver, king-of-the-hill, tips, answer-chaining. The challenge must have restrictions on valid output.
Write a valid submission for the challenge, in a free language found on Wikipedia or esolangs.org or tryitonline. The submission does not have to be competitive, only valid. EDIT: Hashing in your submission is not allowed
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) (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 for the initial idea!

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>

I would ban the question codegolf.stackexchange.com/questions/31695/… — BlueEyedBeast, 2 days ago
@NathanMerrill The current phrasing suggests that if there exists an input that will take an hour to compute the result for, the answer is not valid. This means basically every non-trivial challenge. — quartata, 2 days ago
@NathanMerrill Not feersum here, but as I've commented on Mego's submission, there's no guarantee you won't hit another hash collision before the intended one. I believe cops should have to be certain that their program is indeed the first one that matches, in which case if the cop can verify it then robbers can too. — Sp3000, 2 days ago

DLosc · Answer 1 · 2016-11-18 20:07:10Z

up vote 5 down vote

Pip, 9 bytes

(q`\w+`1)

What it does should be easy to figure out, but the question is why... (TIO)

answered 2 days ago

DLosc

10.1k1453

seems like it's splitting a string on commas and returning the 2nd string in the resulting split O_O I don't know pip, but that's all I got so far. – Yodle 2 days ago

1

@Yodle splitting it into regex words, not just into commas – Maltysen 2 days ago

Translated into pyth I believe this is something like ht:w"\w+"1 / @:w"\w+"1 1, which minifies to htcw / htcz if only whitespace matters, but none of those turn up any results. Also this would be able to pull out minutes from HH:mm:ss format, or extract a variable name from a C-style declaration, or numbers after a decimal point. But I can't find questions which ask for any of those things... :/ – Dave 4 hours ago

add a comment |

Gabriel Benamy · Answer 2 · 2016-11-18 21:19:29Z

up vote 3 down vote

Perl - Cracked by DLosc

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

The code crashes after printing the truthiness, but according to this 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.

edited 2 days ago

answered 2 days ago

Gabriel Benamy

1,4179

4

As if Perl is easier to read than an esolang :P – ETHproductions 2 days ago

1

Yeah I don't know any esolangs, but I also don't know Perl :( – Yodle 2 days ago

Cracked – DLosc 2 days ago

Dangit, I was going to post that exact thing until the post was deleted earlier (I also couldn't figure out how to get it to run on ideone) – ETHproductions 2 days ago

For any future entries, I'll submit an ideone link with an explanation of how to run it. – Gabriel Benamy 2 days ago

add a comment |

Luis Mendo · Answer 3 · 2016-11-18 20:34:56Z

up vote 2 down vote

MATL. Cracked

&:"@FYAYm7>vs

Try it online!

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.

edited 2 days ago

answered 2 days ago

Luis Mendo

33.4k335137

Cracked :-) – ETHproductions 2 days ago

@ETHproductions Oh, how did you find that challenge? I picked it literally randomly .-) – Luis Mendo 2 days ago

add a comment |

Yodle · Answer 4 · 2016-11-18 20:51:46Z

up vote 2 down vote

C#, 590 bytes, Cracked

(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 ,_,

edited 2 days ago

answered 2 days ago

Yodle

1,007419

Lambda is Func<string, int, string>. Looks like it's some keyboard cipher. The number param is how many chars to shift right on the keyboard (ie- ("t",1) = y). – milk 2 days ago

Cracked – milk 2 days ago

Nice, wasn't expecting that to last too long :P – Yodle 2 days ago

add a comment |

Adnan · Answer 5 · 2016-11-18 21:16:19Z

up vote 2 down vote

05AB1E, 27 bytes, cracked!

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!

edited 2 days ago

answered 2 days ago

Adnan

21.9k444167

you post to pastebin?: pastebin.com/QyvpUb9b – tuskiomi 2 days ago

@tuskiomi Hmm, nope that wasn't me. – Adnan 2 days ago

that's strange. I googled your result (you gotta make sure) and that was 1 of 3 results. I wonder there is a bot that's pulling answers?? hmm.. – tuskiomi 2 days ago

Prints the second input – Oliver 2 days ago

1

cracked? – Maltysen 2 days ago

| show 2 more comments

Stewie Griffin · Answer 6 · 2016-11-18 23:40:18Z

up vote 2 down vote

Octave, 68 bytes

@(x,y)find((v=((z=cumsum(x.^2))(y:end)-[0,z(1:end-y)]))==max(v),1)-1

Try it online here.

edited 2 days ago

answered 2 days ago

Stewie Griffin

8,220136107

add a comment |

JHM · Answer 7 · 2016-11-19 06:17:37Z

up vote 2 down vote

아희(Aheui), 0 bytes, Cracked

Try it here! (The program is already typed in for you :p )

edited yesterday

answered 2 days ago

JHM

2,309218

1

Cracked – milk 2 days ago

@milk Wow! That was quick! – JHM 2 days ago

add a comment |

Dada · Answer 8 · 2016-11-19 18:13:43Z

Perl

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.

Martin Ender · Answer 9 · 2016-11-19 22:42:30Z

up vote 2 down vote

Hexagony, 548 bytes

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!

answered yesterday

Martin Ender♦

116k39261608

This challenge is probably not the one you intended, because the output here changes depending on an input number. But without input it prints the string in that challenge, and "robbers need to find any challenge that the submission is a valid submission for". So, cracked? – Luis Mendo 23 hours ago

@LuisMendo it doesn't quite print that string. – Martin Ender♦ 20 hours ago

add a comment |

Mega Man · Answer 10 · 2016-11-20 08:59:46Z

up vote 2 down vote

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...)

edited 19 hours ago

answered yesterday

Mega Man

429213

Now, I don't even understand that code for myself... – Mega Man yesterday

1

It's easy to deobfuscate, but the questions which it might fit all seem to be tagged busy-beaver... – Peter Taylor yesterday

Nice obfuscation! – OldBunny2800 yesterday

This computes 9**99 without printing anything (Also taking a long time to do it) – BlueEyedBeast yesterday

@PeterTaylor It isn't a busy-beaver question, I just took a strange way to solve the question. – Mega Man 19 hours ago

| show 1 more comment

ais523 · Answer 11 · 2016-11-18 20:53:34Z

up vote 1 down vote

Perl, 56 bytes

undef$/;print+(<>^<>)=~y/\x81-\xff\x00-\x80/\x01-\xff/dr

answered 2 days ago

ais523

83516

add a comment |

BlueEyedBeast · Answer 12 · 2016-11-18 22:10:15Z

up vote 1 down vote

Pyke, 42 characters

cFDhkR.dۢ㒶οd-~o@l3{IoKRKn)R+(Jt~o16q*Q|

Try it here!

Subliminal message intended towards cracker.

edited 2 days ago

answered 2 days ago

BlueEyedBeast

13.2k52270

㒶 really belongs there? – Erik the Golfer 2 days ago

yep and so do the other 3 unprintables – BlueEyedBeast 2 days ago

You mean 4 unprintables, right? – Erik the Golfer 2 days ago

yes I do - (one of them is invisible in my browser/blends in) – BlueEyedBeast 2 days ago

add a comment |

Peter Taylor · Answer 13 · 2016-11-19 07:56:23Z

up vote 1 down vote

CJam

{W+W%~1{1$)}{)a1${\(+W%{1$1$-2=>}{+}w}{\;}?)_@*\+~}w+}

This is an anonymous block (function).

answered yesterday

Peter Taylor

33k340117

add a comment |

DLosc · Answer 14 · 2016-11-20 07:11:14Z

up vote 1 down vote

Pip, 13 bytes

V$.C(A*a-A9)a

Try it online (give input as Arguments, not Input).

answered 21 hours ago

DLosc

10.1k1453

add a comment |

Oliver · Answer 15 · 2016-11-21 03:34:36Z

JavaScript, 533 bytes, Cracked! by Dave

_=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.)

Cracked! codegolf.stackexchange.com/a/100591/8927 – Dave 3 hours ago — Dave, 3 hours ago

histocrat · Answer 16 · 2016-11-21 04:14:31Z

up vote 1 down vote

Ruby (cracked by DLosc)

p n = gets.to_i
p n = n*(3*n-1)/2 until n % 7 == 0

edited 19 mins ago

answered yesterday

histocrat

16k42762

@DLosc 0 does output 0. – Martin Ender♦ 18 hours ago

Whoops, I missed the first p when I copied the code over. Cracked, then. – DLosc 2 hours ago

add a comment |

Luis Mendo · Answer 17 · 2016-11-18 20:47:08Z

up vote 0 down vote

MATL

dP7EGn:q^1J2/h)ts_hX=Gs[BE]Wd=~>~GBz*

Try it online!

edited 2 days ago

answered 2 days ago

Luis Mendo

33.4k335137

Try it Online please. – tuskiomi 2 days ago

@tuskiomi Done. But since you have to guess the input format, it's not much use... – Luis Mendo 2 days ago

Numbers seem to work well. – tuskiomi 2 days ago

Cracked! – boboquack yesterday

@boboquack Not quite :-) My code's output is not the Hamming distance of an input (although that may be the case for the inputs you have tried) – Luis Mendo yesterday

| show 2 more comments

BlueEyedBeast · Answer 18 · 2016-11-18 21:23:29Z

up vote 0 down vote

Pyke, 20 bytes

uÄCMX67+*]mA""R`.:

Try it here!

answered 2 days ago

BlueEyedBeast

13.2k52270

add a comment |

Flp.Tkc · Answer 19 · 2016-11-18 21:41:52Z

up vote 0 down vote

Python 3, Cracked!

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)

edited 2 days ago

answered 2 days ago

Flp.Tkc

71919

base64 encoded message: (sum([x%4<1 for x in range(min(D),max(D))])) – BlueEyedBeast 2 days ago

yep, just threw that in for fun @BlueEyedBeast – Flp.Tkc 2 days ago

For some reason. substituting that decoded string in gives me a TypeError >.> – Yodle 2 days ago

@Yodle Does the original code work? If so, just make sure you substitute the decoded string in properly – Flp.Tkc 2 days ago

Yeah it does, which is why I am very confused haha. I'm pretty sure I'm doing it right, just removing base64decode(...).decode() with the string right? – Yodle 2 days ago

| show 4 more comments

Steven H. · Answer 20 · 2016-11-18 22:00:52Z

up vote 0 down vote

Pyth - Cracked by Maltysen

If it helps, Pyth was not any of the answers to the hidden challenge.

u=.+G.*A

Try it out!

edited 2 days ago

answered 2 days ago

Steven H.

1,753117

cracked? – Maltysen 2 days ago

Cracked! That was fast. – Steven H. 2 days ago

add a comment |

BlueEyedBeast · Answer 21 · 2016-11-18 22:07:47Z

Pyke, 3458 bytes

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!

There are a few (30) 0x1b bytes that seem to have been eaten by SE.

Laikoni · Answer 22 · 2016-11-18 22:49:54Z

Haskell

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. "Usage":

Prelude> f "Programming Puzzles & Code Golf"
"rogramming Puzzles "

milk · Answer 23 · 2016-11-19 01:29:47Z

up vote 0 down vote

C#, 91 bytes

_=>{int b=0,w=0;for(;1>w||0<*(_-1);b+=++w**_++<<(9*w));return b%(3<w?903302656:41458688);};

edited 2 days ago

answered 2 days ago

milk

1,3479

add a comment |

BlueEyedBeast · Answer 24 · 2016-11-19 11:42:17Z

up vote 0 down vote

Pyke, 18 bytes

2fm}s"ab"NQlZh.&|!

Try it here!

answered yesterday

BlueEyedBeast

13.2k52270

add a comment |

BlueEyedBeast · Answer 25 · 2016-11-19 12:13:49Z

up vote 0 down vote

Pyke, 19 bytes

4\Y:\m+Ay""%b%y"cl2

Try it here!

answered yesterday

BlueEyedBeast

13.2k52270

add a comment |

JHM · Answer 26 · 2016-11-19 16:42:33Z

up vote 0 down vote

Mathematica, 161 bytes

Which[LetterQ@#,If[UpperCaseQ@#,ToUpperCase,#&][FromLetterNumber~Array~LetterNumber@#],DigitQ@#,Array[IntegerString,FromDigits@#+1,0],True,#]&/@Characters@#<>""&

edited yesterday

answered yesterday

JHM

2,309218

add a comment |

FinW · Answer 27 · 2016-11-19 16:44:06Z

up vote 0 down vote

BrainFuck - 140 Bytes, Cracked by daHugLenny

,>,>,>-[>+<-----]>---[<+>-]<[<<<->>>->+<]>[<<<->>>->+<]>[<<<->>>-]<<<[>+<-]<[>>++++++++++<<-]<[>>>>++++++++++[<++++++++++>-]<<<<-]>>>[>.+<-]

Try It Here!

edited yesterday

answered yesterday

FinW

314

1

Cracked – daHugLenny yesterday

add a comment |

Karl Napf · Answer 28 · 2016-11-19 18:52:43Z

up vote 0 down vote

C++14, Cracked

#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;
}

edited yesterday

answered yesterday

Karl Napf

2,720223

The title says C++14, so you need at least g++ and if your version is not at least 6.2 or something you need -std=c++14 – Karl Napf yesterday

1

Cracked – Peter Taylor yesterday

add a comment |

JHM · Answer 29 · 2016-11-20 00:32:30Z

up vote 0 down vote

Mathematica, 34 bytes, Cracked

±1={±0={}};±n_:=Array[±#&,n,0]

Named function (±).

edited yesterday

answered 2 days ago

JHM

2,309218

I can't tell whether it's this and you overlooked the "No digits 0 to 9 to appear in the code" rule, or whether it's this and you decided to take unary input for obfuscation. – Martin Ender♦ yesterday

Whoops, it's the first one, but I totally missed that rule... I'll change the code to fit the other one :) – JHM yesterday

Cracked then. ;) – Martin Ender♦ yesterday

add a comment |

wat · Answer 30 · 2016-11-21 01:27:00Z

up vote 0 down vote

Ruby, 50 bytes

count = 0; 400.times do count +=1; end; puts count

output: 400

answered 3 hours ago

wat

1,011529

add a comment |

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PPCG Jeopardy: Cops

Uncracked Submissions:

32 Answers 32

Pip, 9 bytes

Perl - Cracked by DLosc

MATL. Cracked

C#, 590 bytes, Cracked

05AB1E, 27 bytes, cracked!

Octave, 68 bytes

아희(Aheui), 0 bytes, Cracked

Perl

Hexagony, 548 bytes

Python, 935 Bytes

Perl, 56 bytes

Pyke, 42 characters

CJam

Pip, 13 bytes

JavaScript, 533 bytes, Cracked! by Dave

Ruby (cracked by DLosc)

MATL

Pyke, 20 bytes

Python 3, Cracked!

Pyth - Cracked by Maltysen

Pyke, 3458 bytes

Haskell

C#, 91 bytes

Pyke, 18 bytes

Pyke, 19 bytes

Mathematica, 161 bytes

BrainFuck - 140 Bytes, Cracked by daHugLenny

C++14, Cracked

Mathematica, 34 bytes, Cracked

Ruby, 50 bytes

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PPCG Jeopardy: Cops

Uncracked Submissions:

32 Answers 32

Pip, 9 bytes

Perl - Cracked by DLosc

MATL. Cracked

C#, 590 bytes, Cracked

05AB1E, 27 bytes, cracked!

Octave, 68 bytes

아희(Aheui), 0 bytes, Cracked

Perl

Hexagony, 548 bytes

Python, 935 Bytes

Perl, 56 bytes

Pyke, 42 characters

CJam

Pip, 13 bytes

JavaScript, 533 bytes, Cracked! by Dave

Ruby (cracked by DLosc)

MATL

Pyke, 20 bytes

Python 3, Cracked!

Pyth - Cracked by Maltysen

Pyke, 3458 bytes

Haskell

C#, 91 bytes

Pyke, 18 bytes

Pyke, 19 bytes

Mathematica, 161 bytes

BrainFuck - 140 Bytes, Cracked by daHugLenny

C++14, Cracked

Mathematica, 34 bytes, Cracked

Ruby, 50 bytes

Your Answer

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