100
\$\begingroup\$

This is the cops' thread. The robbers' thread is here.

Your challenge is to make a program that runs forever without halting1, unless it gets a particular input or inputs2. If it receives that input, it must terminate in a finite amount of time3. This is , so the shortest answer that has not been cracked by a robber within one week of posting wins. After the week has passed, please mark your answer as safe and show the halting input (in a > ! spoiler quote). If a robber cracks your submission, please mark it as cracked and show the halting input (in a > ! spoiler quote).

Submissions are preferred be runnable & crackable on TIO. Submissions not runnable or crackable on TIO are allowed, but please include instructions to download / run them.

Please make your input deterministic, and uniform across all runs. See this meta post for details.

Please, don't "implement RSA" or anything mean to the robbers. Use obscure languages and features, not boring encryption and hashing. I can't enforce this with rules, but you can expect torrential downvotes if all you do is sha(input) === "abcd1234".


1Assuming that the computer doesn't get shut down, break, get engulfed by the sun, overheat in the universe's heat death, or hit the TIO timeout of 60s.

2The program must halt on at least one input. As long as it loops forever on one input and halts on another, it works.

3This must be < 60 seconds, so that the code can be tested on TIO.


Looking for uncracked submissions?

fetch("https://api.stackexchange.com/2.2/questions/135363/answers?order=desc&sort=activity&site=codegolf&filter=!.Fjs-H6J36vlFcdkRGfButLhYEngU&key=kAc8QIHB*IqJDUFcjEF1KA((&pagesize=100").then(x=>x.json()).then(data=>{var res = data.items.filter(i=>!i.body_markdown.toLowerCase().includes("cracked")).map(x=>{const matched = /^ ?##? ?(?:(?:(?:\[|<a href ?= ?".*?">)([^\]]+)(?:\]|<\/a>)(?:[\(\[][a-z0-9/:\.]+[\]\)])?)|([^, ]+)).*[^\d](\d+) ?\[?(?:(?:byte|block|codel)s?)(?:\](?:\(.+\))?)? ?(?:\(?(?!no[nt][ -]competing)\)?)?/gim.exec(x.body_markdown);if(!matched){return;}return {link: x.link, lang: matched[1] || matched[2], owner: x.owner}}).filter(Boolean).forEach(ans=>{var tr = document.createElement("tr");var add = (lang, link)=>{var td = document.createElement("td");var a = document.createElement("a");a.innerHTML = lang;a.href = link;td.appendChild(a);tr.appendChild(td);};add(ans.lang, ans.link);add(ans.owner.display_name, ans.owner.link);document.querySelector("tbody").appendChild(tr);});});
<html><body><h1>Uncracked Submissions</h1><table><thead><tr><th>Language</th><th>Author</th></tr></thead><tbody></tbody></table></body></html>

\$\endgroup\$
9
  • 1
    \$\begingroup\$ @LuisMendo assuming infinite memory is fine \$\endgroup\$
    – user58826
    Jul 24 '17 at 16:29
  • 1
    \$\begingroup\$ @programmer5000 Thanks for clarifying. You should incorporate all this into the challenge text. Answerers are not expected to read all comments \$\endgroup\$
    – Luis Mendo
    Jul 24 '17 at 16:48
  • 7
    \$\begingroup\$ Can we restrict the input to, say, an integer? A robber could pass in some malformed or mistyped input to terminate the program immediately. I'd either have to do lots of careful input validation, or use a language that can catch arbitrary errors. \$\endgroup\$
    – xnor
    Jul 24 '17 at 18:37
  • 3
    \$\begingroup\$ @xnor I think you're supposed to have to do the input checking. \$\endgroup\$
    – Stephen
    Jul 24 '17 at 18:49
  • 1
    \$\begingroup\$ @StepHen That works if the malformed input is still able to be parsed by the interpreter. If I enter an illegal string, R errors immediately, without even entering the function to crack. I would argue that that doesn't count as actually terminating the function. \$\endgroup\$
    – JAD
    Jul 26 '17 at 5:02

115 Answers 115

2
\$\begingroup\$

Python 3.5, 187 bytes Cracked

def f(x):
 t="""def t():
 def t():
  while t:pass
  {}

 t()
t()"""
 if type(x) != str or chr(10) in x:
  while t:pass
 try:t=compile(t.format(x),'m','exec')
 except:
  while t:pass
 exec(t)

Forgot that \r also counts as newline. Fixed Version

\$\endgroup\$
1
2
\$\begingroup\$

JavaScript (Node.js), 36 bytes Cracked

x=>{for(;!(x**x==x&&x!=1&&x+x>x););}

Try it online!

This one should be quite easy ^_^

Solution by Vicente Gallur Valero

1.00000001

My Solutions:

'Infinity' or '-1'

\$\endgroup\$
1
2
\$\begingroup\$

Python 2 (44 characters per TIO): cracked

while 1:
 try:
  x = input()
 except:
  pass

Updated TIO Link to correct for a more trivial example than intended from code golfing:

Try it online!

\$\endgroup\$
6
  • \$\begingroup\$ Dunno if you were trying to golf this or not, but for future whatever, you can switch tabs and spaces for indentation to save bytes (in your case, two spaces -> tabs) and you can remove spaces around =. In case you, or anyone reading, didn't know. \$\endgroup\$
    – Stephen
    Jul 24 '17 at 23:38
  • \$\begingroup\$ @Jonathan Allan: Fixed the pass bug. I'm not sure about the cracked part: I had a different set of inputs in mind that would work from TIO or another thing that did python myapp < myinputs ; can you embed a control-C in a useful fashion in a text file? From linux, almost anything can be terminated by doing control-z and kill -9 %1 so I'm not sure where the dividing line is. \$\endgroup\$
    – Foon
    Jul 24 '17 at 23:39
  • \$\begingroup\$ @StepHen thanks; I hadn't realized this was codegolf until after I posted it and my first attempt to golf it a bit ended up making it trivially easy to break \$\endgroup\$
    – Foon
    Jul 24 '17 at 23:40
  • \$\begingroup\$ Let's say "not cracked" and wait for a better one then! ...For readers - I am retracting exit() <newline> <ctrl>C (or *nix equivalent) as per this answer in light of a better one... \$\endgroup\$ Jul 24 '17 at 23:41
  • 2
    \$\begingroup\$ Cracked. The "possibly" was irking me. \$\endgroup\$
    – Veedrac
    Jul 25 '17 at 3:08
2
\$\begingroup\$

R, 37 bytes, Cracked

f=function(x){eval(x);x=T;while(x){}}

Try it online!

Intended solution (basically the same as the cracked version):

x=T<-F

\$\endgroup\$
1
2
\$\begingroup\$

Python 3, 45 bytes (cracked, and again!)

A bit of an odd one, and probably easier than the last, but it's shorter.

def T(x,t=type):
 while t(x)is t or T(t(x)):0

Pass one argument to T by appending code to this program. That argument may be initialized imperatively. This answer is the kind of thing I'm looking for.

\$\endgroup\$
8
  • \$\begingroup\$ This seems easily crackable by specifying a second argument. \$\endgroup\$
    – Dennis
    Jul 25 '17 at 5:05
  • 1
    \$\begingroup\$ @Dennis It's a one argument function. *looks both ways*. I wouldn't count such an answer in the spirit of the question. Passing 12 arguments halts too. \$\endgroup\$
    – Veedrac
    Jul 25 '17 at 9:16
  • 2
    \$\begingroup\$ Cracked \$\endgroup\$ Jul 25 '17 at 16:08
  • 2
    \$\begingroup\$ Or was this what you're looking for? \$\endgroup\$
    – ecatmur
    Jul 25 '17 at 16:14
  • 1
    \$\begingroup\$ I have to admit, I would never have come up with that answer myself without looking in the CPython test suite \$\endgroup\$
    – ecatmur
    Jul 25 '17 at 16:21
2
\$\begingroup\$

JS (ES6), 23 bytes (Cracked)

x=>{while(!x||x>="\n");}

Fairly easy, but not that easy.

-1 byte thanks to Step Hen, -3 thanks to a mistake pointed out by Step Hen

\$\endgroup\$
6
  • \$\begingroup\$ @StepHen You can't assume input is an a varible, but lambda functions like this are allowed. While is a block statement, meaning it has to be wrapped in {} when in an arrow functions. \$\endgroup\$
    – user58826
    Jul 24 '17 at 16:18
  • 1
    \$\begingroup\$ @StepHen oh oops! \$\endgroup\$
    – user58826
    Jul 24 '17 at 16:21
  • 2
    \$\begingroup\$ That confused the crap out of me. Cracked \$\endgroup\$
    – Stephen
    Jul 24 '17 at 16:25
  • 2
    \$\begingroup\$ Cracked \$\endgroup\$
    – Shaggy
    Jul 24 '17 at 16:26
  • \$\begingroup\$ Am I missing something or does the Braingolf solution work here too? \$\endgroup\$
    – Neil
    Jul 24 '17 at 16:54
2
\$\begingroup\$

JS (ES6), 81 75 bytes (Cracked)

x=>eval("for(var i=0;i<1/0;i+="+(x.slice?Math.abs(Number(x))+50||1:1)+");")

Try it online!

\$\endgroup\$
1
  • \$\begingroup\$ Cracked \$\endgroup\$
    – Stephen
    Jul 24 '17 at 18:06
2
\$\begingroup\$

C#, 63 bytes (Cracked)

a=>{while(1>0)if(string.IsInterned(new string(a))!=null)break;}

Compiles to a Action<char[]>.

Try it online!

An empty array and null all seem to work. This is because string.Empty, or "", is always interned.

\$\endgroup\$
6
  • \$\begingroup\$ Hmm, based on my knowledge of IsInterned, it returns true for an Interned string, which is a string that is a string literal within the code, or has been interned with Intern, this means you're abusing some undocumented behaviour of IsInterned somehow, because there are no string literals defined. \$\endgroup\$
    – Mayube
    Jul 26 '17 at 9:27
  • \$\begingroup\$ @Mayube It returns the string not a bool which I found odd myself. I suppose it isn't as intuitive as you'd think it would be though. \$\endgroup\$ Jul 26 '17 at 9:37
  • \$\begingroup\$ that's what I meant sorry, IsInterned()!=null returns true if the string is Interned, and false otherwise. \$\endgroup\$
    – Mayube
    Jul 26 '17 at 9:41
  • \$\begingroup\$ @Mayube It's been cracked, I thought this one was harder than that. \$\endgroup\$ Jul 26 '17 at 9:45
  • \$\begingroup\$ Aaah, my problem is I was thinking input from STDIN, not just argument to the method :P \$\endgroup\$
    – Mayube
    Jul 26 '17 at 9:45
2
\$\begingroup\$

JavaScript (Node.js), 52 46 bytes (Cracked)

x=>{try{(!(x*1)||x*1)&&0()}catch(e){for(;;);}}

Try it online!

Intended:

// break excluded middle:
{valueOf:()=>this.q?0:this.q=1}

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Cracked? \$\endgroup\$
    – tsh
    Jul 26 '17 at 1:53
  • \$\begingroup\$ Cracked in a less clever way! \$\endgroup\$ Jul 26 '17 at 10:59
  • \$\begingroup\$ @DomHastings that's just mean \$\endgroup\$
    – Alex Varga
    Jul 26 '17 at 15:13
  • \$\begingroup\$ Yeah, I thought so! 😊 Enjoyed playing with it though, thanks! \$\endgroup\$ Jul 26 '17 at 18:32
2
\$\begingroup\$

CPython 2.7, 96 93 100 bytes Cracked

There are so many loopholes in python! It is really hard to write code which can't be manipulated or produce vastly unexpected results.

def f(x):
 try:
  s=str(x)
  [c for c in s if c not in'x.']and 0/0
  exec(s)
 finally:
  while(1):1

Not too hard, I'm afraid.

\$\endgroup\$
11
  • \$\begingroup\$ I'm new to code golf so I am not sure if I need to make this a function or if stand alone code like this is OK. Can anyone point me at some rule or clarify this for me? \$\endgroup\$
    – Sudo Bash
    Jul 26 '17 at 21:42
  • \$\begingroup\$ on my machine, del str immediately throws NameError regardless of the input I give. \$\endgroup\$
    – tbodt
    Jul 26 '17 at 21:56
  • \$\begingroup\$ @tbodt So it gets caught in the while loop at the end right? I wasn't sure if the rules of this challenge allows the robbers to override str, so I did del str just in case. So if you don't override it, you get caught in the loop. I would ask the OP but I don't have comment priv yet. \$\endgroup\$
    – Sudo Bash
    Jul 26 '17 at 21:58
  • \$\begingroup\$ If you write your program as a function, the input would come from function arguments, otherwise it would come from outside the program (e.g. input() or sys.argv) \$\endgroup\$
    – tbodt
    Jul 26 '17 at 22:00
  • \$\begingroup\$ OK, thanks. I made it a function and removed del str because I assume that it would be cheating to override str since that is not technically input. \$\endgroup\$
    – Sudo Bash
    Jul 26 '17 at 22:06
2
\$\begingroup\$

Pyth, 5 bytes, Cracked, Cracked

##.vz

Code

I believe there's (essentially) only a single solution.

\$\endgroup\$
2
  • \$\begingroup\$ Cracked? \$\endgroup\$
    – Adnan
    Jul 26 '17 at 11:37
  • \$\begingroup\$ Cracked. \$\endgroup\$
    – Veedrac
    Jul 26 '17 at 11:37
2
\$\begingroup\$

C++ (gcc/MSVC), 62 bytes, cracked

template<class T,T=T{}>T*f(T t){for(static T*a;!a;(T&)a=t=0);}

Tested on Linux and Windows, with and without optimizations enabled. Under MSVC, you may wish to use #pragma warning(default:4716).

class T{}T::*t;

A null pointer-to-data-member is (typically, on modern platforms) assigned a numeric value of -1, as opposed to other null pointers, which are assigned a numeric value of 0. This is so that the pointer-to-data-member can store the offset of the data member from the beginning of the class, since this offset could be 0 for the first data member.

Try it online!

\$\endgroup\$
2
  • \$\begingroup\$ Cracked (Anyone add an explanation for me? I don't know how that works.) \$\endgroup\$
    – DELETE_ME
    Jul 27 '17 at 11:08
  • \$\begingroup\$ @user202729 Excellent! Added an explanation in the spoiler. \$\endgroup\$
    – ecatmur
    Jul 27 '17 at 12:45
2
\$\begingroup\$

Python 3.5, 203 bytes (Cracked with Intended Solution and Cracked)

def f(x):
 t="""def t():
 def t():
  while t:pass
  {}

 t()
t()"""
 import re
 if type(x) != str or re.search(r"\s",x,re.UNICODE):
  while t:pass
 try:t=compile(t.format(x),'m','exec')
 except:
  while t:pass
 exec(t)

There is actually a third way to crack it

My Answer:

f("t=t")
Python decides at compile time if a variable is local or not by checking if it is assigned anywhere in the function
So even if the statement is never executed it still changes the behaviour of the function(in this case causing an UnboundLocalError because it is never assigned)

\$\endgroup\$
2
  • \$\begingroup\$ I won't resubmit @JonathanAllan cracked version just by changing '\r' to chr(12), but you might want to check for that also. \$\endgroup\$ Jul 25 '17 at 15:48
  • \$\begingroup\$ Cracked \$\endgroup\$ Jul 25 '17 at 21:23
2
\$\begingroup\$

Node.js, 129 bytes Cracked

q=>{try{v=require("vm");c=v.createContext();new v.Script(String.prototype.toString.apply(q)).runInContext(c)}catch(a){}for(;;){}}

Unwanted: tampering with require/vm (eg require=1;q())

\$\endgroup\$
2
  • \$\begingroup\$ which version of Node? \$\endgroup\$ Jul 27 '17 at 23:56
  • \$\begingroup\$ Cracked \$\endgroup\$ Jul 28 '17 at 0:44
2
\$\begingroup\$

Swift 3, 43 bytes, Cracked

func f(i:Int){let p=0;while i==p||i != p{}}

Call with f(i: /*your input here*/)

More readable:

func gandalf(input: Int) {
    let password = 0
    while input == password || input != password {
        print("You shall not pass!")
    }
    
    print("nvm")
}

Call with gandalf(input: /*your input here*/)

Try it here!

I affirm that I have a working crack for this.

\$\endgroup\$
10
  • \$\begingroup\$ The code at the "Try it here!" link does not match the one in the post. I don't know whether it matters - I have no idea how swift works. \$\endgroup\$
    – NieDzejkob
    Jul 26 '17 at 18:25
  • \$\begingroup\$ @NieDzejkob Yes it does, it matches the ungolfed code (for the robbers' convenience, of course :P) \$\endgroup\$
    – Alexander
    Jul 26 '17 at 18:37
  • \$\begingroup\$ @Alexander It doesn't; password is inlined. \$\endgroup\$
    – Veedrac
    Jul 26 '17 at 18:42
  • \$\begingroup\$ @Veedrac Oh, true! \$\endgroup\$
    – Alexander
    Jul 26 '17 at 18:47
  • \$\begingroup\$ Can the crack be a runtime error? If so, cracked. I am guessing this is not what you had in mind, though. \$\endgroup\$
    – Daniel
    Jul 27 '17 at 21:07
2
\$\begingroup\$

Python 3, 35 bytes, cracked

def f(o):
 while bool(o)==bool(o):0

Try it online!

Not too hard...

Intended solution (ungolfed, 'cause screw it):

class LOL: def __init__(self): self.lol = True def __bool__(self): self.lol = not self.lol return self.lol lol = LOL() f(lol)

\$\endgroup\$
1
2
\$\begingroup\$

Python, 91 bytes, Cracked

x=input()
try:y=len(x)<10and'*'not in x and len(str(eval(x)))==307
except:y=0
while not y:0

raw_input for Python 2.

No one found the intended answer but the solutions found were very creative and impressive.

A hint for the intended answer is:

it consists of only letters.

The intended answer is:

copyright

\$\endgroup\$
10
  • \$\begingroup\$ Cracked. \$\endgroup\$
    – Sisyphus
    Jul 29 '17 at 12:31
  • \$\begingroup\$ @Sisyphus Very clever. I've shortened the length limit so that won't work any more. \$\endgroup\$
    – Alex Hall
    Jul 29 '17 at 12:38
  • \$\begingroup\$ Probably not what you had in mind. Still, it's cracked on my system. \$\endgroup\$ Jul 29 '17 at 15:02
  • \$\begingroup\$ @EricDuminil nice, and a surprising coincidence, but not what I had in mind. I've changed the length limit again. This is the real one now, I promise! Also the intended answer does work on both python 2 and 3 (surprisingly) at least on my system. \$\endgroup\$
    – Alex Hall
    Jul 29 '17 at 15:37
  • \$\begingroup\$ Cracked.. I had another answer too that did fit in the original length limit, but not in the edited one: "_".__mul__(307). \$\endgroup\$
    – hvd
    Jul 30 '17 at 14:45
2
\$\begingroup\$

JavaScript ES6 (Node.js >= 6.0), 326 bytes (Safe)

((v,e=require('events').EventEmitter.prototype,l=_=>{while(1);},P=Proxy,p=process)=>(p.kill=p.exit,e.listeners=new P(e.listeners,{apply:(t,c,a)=>a[0]=='exit'?[]:t.apply(c,a)}),e.removeAllListeners=new P(e.removeAllListeners,{apply:(t,c,a)=>a[0]=='exit'?c:t.apply(c,a)}),p.on('exit',l),i=>{try{v(i)}catch(o){}l()}))(s=>eval(s))

Much more simplified than my other challenge was, I'm no longer attempting to obfuscate the approach for the solution, which saved me a lot of bytes. This time around, process.kill() won't work.

"delete process._events.exit;process.exit()"

Try it online!

\$\endgroup\$
2
  • \$\begingroup\$ Woah! Nice job! \$\endgroup\$
    – user58826
    Aug 7 '17 at 19:09
  • \$\begingroup\$ It looks like you're the winner, although I was expecting a much shorter code hehe \$\endgroup\$
    – Piyin
    Aug 8 '17 at 15:43
2
\$\begingroup\$

#Batch, 26 24 28 bytes - Cracked(Unintened)

Unfortunately, TIO doesn't support batch, so you will need to try on a command prompt.

:L
set/pp=
echo("%p%"
goto l

The unintended solution:

"&&exit 
(Note the trailing space!) This undoes the quotes in echo(" and execute the exit command. Actually, the exit command can be changed to lots of other commands that causes an exit.

The intended solution:

Hint: The intended solution does not use &&, but uses variable expansion bug.

\$\endgroup\$
11
  • \$\begingroup\$ You should not reveal the inputs it crashes on... That's the whole point of the challenge. \$\endgroup\$
    – Mr. Xcoder
    Jul 29 '17 at 9:28
  • \$\begingroup\$ @Mr.Xcoder Change is in. \$\endgroup\$
    – stevefestl
    Jul 29 '17 at 9:29
  • \$\begingroup\$ @Mr.Xcoder Oh well. I shouldn't change the code. \$\endgroup\$
    – stevefestl
    Jul 29 '17 at 9:33
  • \$\begingroup\$ I deleted my answer. \$\endgroup\$
    – Mr. Xcoder
    Jul 29 '17 at 9:34
  • \$\begingroup\$ Nevermind, I'll let someone else crack it. \$\endgroup\$
    – Mr. Xcoder
    Jul 29 '17 at 9:35
2
\$\begingroup\$

Python 2, 86 bytes (safe)

def _(x):exec x
try:(input()or _("while 1:0"))and _("while 1:0")
except:_("while 1:0")

I know this is a late answer, but I thought of what I think is an interesting way to do this.

Update and explanation:

The answer is

_("__import__('os')._exit(0)"). In Python 2, the input function evals its argument before returning the resulting value. In order to bypass the infinite loops, the input string must somehow lead input to exit the program without raising an exception. Ordinarily, this should be impossible, because eval cannot execute arbitrary code by itself in Python 2. However, the _ function may be called by a string passed to eval, and since _ is itself a wrapper for the exec statement, which can execute arbitrary code, this input is able to function by calling _ with a string containing a call to the os._exit function, which exits the program immediately without erroring.

Actually, I just noticed that the _ exec wrapper isn't strictly necessary, as __import__('os')._exit(0) is technically an expression rather than a statement and thus can be passed to eval without a syntax error.

Regardless, one should never use the input function in Python 2, especially in conjunction with a wrapper for exec. Use raw_input instead and then convert the value to whatever type you want.

\$\endgroup\$
2
\$\begingroup\$

05AB1E - 4 bytes

[I#]

The answer is easy if you know 05AB1E

\$\endgroup\$
1
2
\$\begingroup\$

JavaScript (Node.js), 108 bytes, Double Cracked

f=y=>(x=>{for(;typeof y=='string'&&x===x;);})(f['constructor'](`return +[${y.replace(/[^W-`*-0]/g,'')}]`)())

Try it online!

Fixed a bug that allowed errors...

My intended solution was [][+[]]+[]][+[]. This evaluates to coercing undefined to an integer, which is NaN, and NaN !== NaN.

\$\endgroup\$
10
  • \$\begingroup\$ It seems like for an input like 'a' it just errors and halts. Am I doing anything wrong? \$\endgroup\$ May 21 at 13:13
  • \$\begingroup\$ @CommandMaster Slaps face It should work now. \$\endgroup\$
    – emanresu A
    May 21 at 21:07
  • \$\begingroup\$ Probably not what you meant, but cracked. Perhaps making sure y is treated as a string would make this trickier? \$\endgroup\$
    – Etheryte
    May 21 at 21:58
  • \$\begingroup\$ Cracked, although I am guessing your intended solution was more clever 😏 (edit: ninja'd) \$\endgroup\$
    – EasyasPi
    May 21 at 21:59
  • \$\begingroup\$ @EasyasPi Added an explanation. \$\endgroup\$
    – emanresu A
    May 21 at 22:03
1
\$\begingroup\$

JavaScript (Node.js), 35 bytes (Cracked and Cracked)

I hope that I didn't miss anything ¯\_(ツ)_/¯:

x=>y=>{while(!(x>=y&&x<=y&&x!=y));}

Try it online!

Intended solution:

{}, {}

\$\endgroup\$
2
  • \$\begingroup\$ Cracked. Nice job! \$\endgroup\$
    – user58826
    Jul 24 '17 at 18:02
  • \$\begingroup\$ Cracked? \$\endgroup\$ Jul 24 '17 at 18:05
1
\$\begingroup\$

Add++, 106 bytes, Cracked

+?
-1
=?
-1
=?
N
W,+0
+?
+?
+?
>?
-?
V
V
^2
V
=?
N
W,+0
+G
-G
V
=?
N
W,+0
+G
*G
-?
W,+0
+?
>
-24
-?
W,+0
O

Try it online!

This is really mean, to anyone who tries to crack it. So, uh, have fun! I forgot about empty input!

\$\endgroup\$
2
  • \$\begingroup\$ Cracked \$\endgroup\$
    – Mr. Xcoder
    Jul 24 '17 at 18:14
  • \$\begingroup\$ Uh, it doesn't go into an infinite loop as it is (no input). \$\endgroup\$ Jul 24 '17 at 18:14
1
\$\begingroup\$

MATL, 7 bytes (cracked and cracked)

jU`ttX=

Try it online!

\$\endgroup\$
5
  • \$\begingroup\$ Cracked, the empty input errors and terminates the program. \$\endgroup\$
    – Mr. Xcoder
    Jul 24 '17 at 16:53
  • \$\begingroup\$ @Mr.Xcoder Aw. That was not the intended solution, but well done! \$\endgroup\$
    – Luis Mendo
    Jul 24 '17 at 17:14
  • \$\begingroup\$ Perhaps I'll find the intended one too \$\endgroup\$
    – Mr. Xcoder
    Jul 24 '17 at 17:14
  • \$\begingroup\$ Cracked \$\endgroup\$
    – PunPun1000
    Jul 24 '17 at 17:51
  • \$\begingroup\$ @PunPun1000 That was the intended solution :-) Well done!! N also works (it directly means NaN) \$\endgroup\$
    – Luis Mendo
    Jul 24 '17 at 18:12
1
\$\begingroup\$

JavaScript (ES6), 28834 bytes, Cracked

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Try it online!

Answer is 991, checks for input * 7 == 6937

\$\endgroup\$
3
  • \$\begingroup\$ Cracked \$\endgroup\$
    – user58826
    Jul 24 '17 at 19:22
  • \$\begingroup\$ Should you mention JSFuck in your heading? \$\endgroup\$
    – kamoroso94
    Jul 25 '17 at 12:22
  • \$\begingroup\$ @kamoroso94 JSF is a subset of JavaScript, it's perfectly valid JavaScript, so not really - it's just a (popular) method of obfuscating JavaScript that has a name. \$\endgroup\$
    – Stephen
    Jul 25 '17 at 12:32
1
\$\begingroup\$

Octave, 25 bytes (Cracked and Cracked)

x=input('');while 1|x,end

Try it online!

\$\endgroup\$
2
1
\$\begingroup\$

RProgN 2, 11 bytes Cracked

` .iS‘"{".C

Try it online!

Cracks:

# A single # will cause the { to be treated as a function literal, rather than executing it, which avoids the loop.

\$\endgroup\$
3
  • \$\begingroup\$ Input through arguments? \$\endgroup\$ Jul 24 '17 at 23:06
  • \$\begingroup\$ Yep. @ConorO'Brien \$\endgroup\$
    – ATaco
    Jul 24 '17 at 23:07
  • \$\begingroup\$ Cracked. \$\endgroup\$ Jul 24 '17 at 23:11
1
\$\begingroup\$

JavaScript (in Node.js), 65 bytes Cracked

x=>{try{for(;!x||x!=0||x instanceof Object;);}catch(e){for(;;);}}

Try it online!

'0'

Let's try another one: https://codegolf.stackexchange.com/a/135562/44718

\$\endgroup\$
4
  • \$\begingroup\$ Cracked \$\endgroup\$ Jul 25 '17 at 2:17
  • 1
    \$\begingroup\$ Beat me by one minute. \$\endgroup\$ Jul 25 '17 at 2:19
  • 2
    \$\begingroup\$ @HaumedRahmani aha, not what i want, maybe i will post another cops.... T_T \$\endgroup\$
    – tsh
    Jul 25 '17 at 2:21
  • \$\begingroup\$ @HaumedRahmani OK, just a mirror edit... codegolf.stackexchange.com/a/135562/44718 \$\endgroup\$
    – tsh
    Jul 25 '17 at 2:25
1
\$\begingroup\$

J, 22 bytes, Cracked

$:@#~^:(0([=[e.{#])#~)

This will run forever assuming memory was unlimited, except when given a specific type of input.

\$\endgroup\$
5
  • \$\begingroup\$ Cracked. \$\endgroup\$ Jul 25 '17 at 2:46
  • \$\begingroup\$ You might want to suggest the user to test on TIO, this crashed my computer (at least on the windows version) \$\endgroup\$ Jul 25 '17 at 2:51
  • \$\begingroup\$ @ConorO'Brien It continuously expands the array if it isn't the right input. It will continue to use more memory, and at least on my (Linux) machine, the process gets killed for using too much memory. \$\endgroup\$
    – miles
    Jul 25 '17 at 2:54
  • \$\begingroup\$ Windows isn't that nice. (Also, the conditional fails for non-positive numbers and floats and infinity as well, all of which error) \$\endgroup\$ Jul 25 '17 at 3:00
  • \$\begingroup\$ I think I found the intended solution and have edited into my answer \$\endgroup\$ Jul 25 '17 at 3:03

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