Will it halt? (Robbers)

Question

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

Your challenge is to crack a cop's submission by finding an input that makes it halt. You don't have to find out why, or all inputs that make it halt if there's more than one, or the input the cop intended, just one input will do.

Once you have cracked a submission, post a link to it in a comment or edit to the cop's post. You can also flag your submission for a mod to edit it in the cop's post. Also, post the input used and a link to the cop's post in an answer in this thread. The robber that cracks the most submissions wins.

Multiple people can post cracks to the same cop submission, as long as they are different.

(If SE converted your duplicate answer to a comment, you may want to vote on this feature request)

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.innerText = 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>

Does different mean different inputs (say, all inputs ending with 2 crack the cop's post - can you different people post different numbers ending in 2?) or different families of inputs, or different types of inputs? — Stephen, Jul 24, 2017 at 17:03
Multiple people can post cracks to the same cop submission... Please define different. — Dennis, Jul 25, 2017 at 13:41

Cristian Lupascu · Accepted Answer · 2017-07-25 14:46:58Z

1

JavaScript, Step Hen

One input that stops the program is the object literal {valueOf: () => false}:

var f=x=>{while(!(x&&x==false));};
f({valueOf: () => false});
console.log('Done!');

answered Jul 25, 2017 at 14:46

Cristian Lupascu

8,6792 gold badges25 silver badges52 bronze badges

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Dennis · Accepted Answer · 2017-07-25 14:53:29Z

1

JavaScript, Step Hen

[]

Try it online!

answered Jul 25, 2017 at 14:53

Dennis

210k41 gold badges370 silver badges823 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Python 3 (CPython), Veedrac

import gc
gc.set_debug(gc.DEBUG_SAVEALL)
gc.collect()
l(gc.garbage[0])

Try it online!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jul 25, 2017 at 14:59

ecatmur

1,68510 silver badges15 bronze badges

\$\begingroup\$ One man's garbage... (nicely done) \$\endgroup\$
– Veedrac
Jul 25, 2017 at 15:05

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score 1 · Accepted Answer · 2017-07-25 20:06:01Z

1

JS, Step Hen

"void 0"

Simple: the 5th char is , the string doesn't have any trailing or leading whitespace, and evaled results in undefined.

edited Jul 25, 2017 at 20:06

answered Jul 25, 2017 at 15:21

user58826

1

\$\begingroup\$ How did you know about that :( good job \$\endgroup\$
– Stephen
Jul 25, 2017 at 15:22
1

\$\begingroup\$ @programmer5000 hyperboleandahalf.blogspot.com/2010/04/… \$\endgroup\$
– Maya
Jul 25, 2017 at 18:32

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

R Jarko Dubbeldam

f=function(x)while(1)if(length(x))grep(x,'')

Fails with a function argument, e.g.,

f(mean)
# Error in as.character(pattern) : 
  cannot coerce type 'closure' to vector of type 'character'

Try it online!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jul 25, 2017 at 21:22

Gregor Thomas

2931 silver badge7 bronze badges

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Veedrac · Accepted Answer · 2017-07-26 03:25:07Z

1

anyfix, Mayube

-5

Easy enough once you learn the neat language. Basically once you make 0 from the addition, the rest is fluff.

Try it online.

answered Jul 26, 2017 at 3:25

Veedrac

6313 silver badges7 bronze badges

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tsh · Accepted Answer · 2017-07-26 05:37:10Z

1

JavaScript by w0lf

Object(NaN)

Another crack, more elegant way.

answered Jul 26, 2017 at 5:37

tsh

34.7k2 gold badges33 silver badges126 bronze badges

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Veedrac · Accepted Answer · 2017-07-26 11:42:26Z

1

R, Jarko Dubbeldam

expression(quit())

Try it online.

answered Jul 26, 2017 at 11:42

Veedrac

6313 silver badges7 bronze badges

\$\begingroup\$ Nice, indeed the intended solution. \$\endgroup\$
– JAD
Jul 26, 2017 at 11:44

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

CPython 3.6, by wizzwizz4

lambda *a,main=__import__("__main__"):(exec("""main.bool=lambda:True"""),1/0)

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jul 26, 2017 at 15:48

The Fifth Marshal

6,2471 gold badge24 silver badges44 bronze badges

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Dennis · Accepted Answer · 2017-07-26 16:29:59Z

1

C++, SIGSEGV

Segfaults.

Try it online!

answered Jul 26, 2017 at 16:29

Dennis

210k41 gold badges370 silver badges823 bronze badges

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Veedrac · Accepted Answer · 2017-07-26 18:40:07Z

1

C++, ecatmur

Truly the most beautiful code I have ever written.

#include <iostream>

auto killmenow = 0;

struct Tee {
    Tee() { if (killmenow++ > 10) { exit(0); } }
};

auto t = Tee{};

int main() {
    f<Tee &, t>(t);
}

This question really confuses me, ergo the above disaster.

Try it online.

answered Jul 26, 2017 at 18:40

Veedrac

6313 silver badges7 bronze badges

\$\begingroup\$ Totally not the intended solution, but absolutely a valid crack. \$\endgroup\$
– ecatmur
Jul 26, 2017 at 19:52

Add a comment |

tbodt · Accepted Answer · 2017-07-26 23:37:54Z

1

Python, Sudo Bash

import signal

class x:
    @property
    def x(self):
        signal.alarm(1)
    def __str__(self):
        return 'x.x'

answered Jul 26, 2017 at 23:37

tbodt

2,4363 gold badges15 silver badges26 bronze badges

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2 revs, 2 users 97% · Accepted Answer · 2017-07-27 00:04:50Z

Python 3, Magenta

def f(x):
 y=str(x)
 while not(x and(y[-1]+y[:-1])==str(2*x)):a=1

The trick is to find a number x which when multiplied by two is equal to itself when all the digits are shifted to the right, with wrapping.

Isaacg found the first number for which this works (105263157894736842), and using the same pattern, I found a bunch more inputs: 210526315789473684, 421052631578947368, 315789473684210526 all work.

Because I didn't make the original crack, and @Isaacg hasnt posted this yet, I made this as CW.

Note that these numbers are formed from the decimal expansion of 1/19.

dhaffey · Accepted Answer · 2017-07-27 04:10:12Z

1

Python 3.5, Siphor

'yield'

TIO

answered Jul 27, 2017 at 4:10

dhaffey

1111 bronze badge

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Daniel · Accepted Answer · 2017-07-28 07:00:49Z

1

Swift, Alexander

func ==(lhs: Int, rhs: Int) -> Bool {
    return false
}

Redefine the == operator to always return false. This means that now, 0 works because 0 == 0 is now false, and 0 != 0 is still false

Alternatively, redefine the != operator to always return false so that every integer other than 0 will halt.

answered Jul 28, 2017 at 7:00

Daniel

6,7191 gold badge20 silver badges58 bronze badges

\$\begingroup\$ Solved, but I have another! It's shorter, too! codegolf.stackexchange.com/a/136343/11490 \$\endgroup\$
– Alexander
Jul 28, 2017 at 8:02
\$\begingroup\$ I'm just blown away that this is something you can do. First I've seen a language allow the redefinition of critical operators. \$\endgroup\$
– Allison
Aug 2, 2017 at 5:40

Add a comment |

hyper-neutrino · Accepted Answer · 2017-07-29 00:24:55Z

1

Python 3: totallyhuman's post, 143 bytes

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

# My code starts here

class A:
	def __init__(s):s.a=False
	def __bool__(s):s.a=not s.a;return s.a

f(A())

Try it online!

answered Jul 29, 2017 at 0:24

hyper-neutrino♦

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Sisyphus · Accepted Answer · 2017-07-29 12:15:54Z

1

Python 2, Alex Hall

__builtin__,True,

Try it online!

We override True to the empty string, which is falsy.

answered Jul 29, 2017 at 12:15

Sisyphus

14.8k3 gold badges45 silver badges86 bronze badges

\$\begingroup\$ Well done. That was the intended solution. \$\endgroup\$
– Alex Hall
Jul 29, 2017 at 12:18

Add a comment |

Sisyphus · Accepted Answer · 2017-07-29 12:31:12Z

1

Python 3, Alex Hall

eval(input())
__import__("os")._exit(0)

Try it online!

Clearly not the intended answer. Bypasses the length limit by evaluating the input.

answered Jul 29, 2017 at 12:31

Sisyphus

14.8k3 gold badges45 silver badges86 bronze badges

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PunPun1000 · Accepted Answer · 2017-07-29 18:18:40Z

1

C (gcc), dj0wns

%s12345678

Try it Online!

Since c is only defined as an 8 ('\b') character array and the fgets is passed a second argument of 12 ('\f') longer strings begin to corrupt d which is located after c on the stack. Adding enough characters will cause the snprintf to segfault when it tries to reference the pointer d which has been corrupted

answered Jul 29, 2017 at 18:18

PunPun1000

9834 silver badges7 bronze badges

\$\begingroup\$ Yay! You found it! \$\endgroup\$
– dj0wns
Jul 30, 2017 at 18:48
\$\begingroup\$ I wanted to find a way to make a buffer overflow non-obvious so my misdirection added alternate solutions sadly. \$\endgroup\$
– dj0wns
Jul 30, 2017 at 18:59

Add a comment |

hvd · Accepted Answer · 2017-07-30 14:44:30Z

1

Python, Alex Hall

"%307d"%1

This creates a truthy string consisting of 306 spaces followed by 1.

answered Jul 30, 2017 at 14:44

hvd

3,6641 gold badge15 silver badges17 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Python, by veedrac

import sys;sys.settrace(-1)

The -1 can be replaced with any other non-callable object

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jul 30, 2017 at 21:22

The Fifth Marshal

6,2471 gold badge24 silver badges44 bronze badges

\$\begingroup\$ Well, that was fast. \$\endgroup\$
– Veedrac
Jul 30, 2017 at 23:10
\$\begingroup\$ Was this the intended solution? \$\endgroup\$
– The Fifth Marshal
Jul 30, 2017 at 23:13
\$\begingroup\$ Yep, this was it. \$\endgroup\$
– Veedrac
Jul 30, 2017 at 23:15
\$\begingroup\$ I've done far worse things with sys.setprofile and sys.settrace \$\endgroup\$
– The Fifth Marshal
Jul 30, 2017 at 23:16

Add a comment |

DELETE_ME · Accepted Answer · 2017-07-31 05:51:33Z

1

Batch, SteveFest

Just undo the quote.

"&&exit

(with a trailing )

Another unintended solution:

"|exit

edited Jul 31, 2017 at 5:51

answered Jul 29, 2017 at 12:43

DELETE_ME

17.3k2 gold badges36 silver badges70 bronze badges

\$\begingroup\$ Well done! Although this is not me intended solution \$\endgroup\$
– stevefestl
Jul 29, 2017 at 12:45
\$\begingroup\$ I will add the link to here shortly \$\endgroup\$
– stevefestl
Jul 29, 2017 at 12:45

Add a comment |

Leaky Nun · Accepted Answer · 2017-08-01 14:17:10Z

1

Python 3, Alex Hall

1<<1017

Try it online!

answered Aug 1, 2017 at 14:17

Leaky Nun

49.7k6 gold badges108 silver badges282 bronze badges

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Taylor Raine · Accepted Answer · 2017-10-03 20:12:46Z

1

JS, Step Hen

Lots of obfuscation!

Try it online!

edited Oct 3, 2017 at 20:12

Taylor Raine

8,3662 gold badges25 silver badges48 bronze badges

answered Jul 24, 2017 at 19:22

user58826

1

\$\begingroup\$ How did you unobfuscate this? \$\endgroup\$
– user41805
Jul 24, 2017 at 19:25
\$\begingroup\$ @Cowsquack Removed everything except the obfuscated code, than removed the (). \$\endgroup\$
– user58826
Jul 24, 2017 at 19:31
\$\begingroup\$ * Step Hen :P and @Cowsquack JSF*** decoder \$\endgroup\$
– Stephen
Jul 24, 2017 at 19:35

Add a comment |

Dennis · Accepted Answer · 2018-06-09 14:05:56Z

1

Python 3 by Veedrac

A single 0xff byte for UTF-8, or anything else that cannot be decoded by the selected encoding.

Try it online!

edited Jun 9, 2018 at 14:05

answered Jul 27, 2017 at 3:35

Dennis

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

05AB1E, MilkyWay90

TIO

Pretty easy one, all I had to do was look up the command list for the language.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 15, 2018 at 0:17

Draco18s no longer trusts SE

3,38914 silver badges25 bronze badges

\$\begingroup\$ Yeah, my one was easy to crack. \$\endgroup\$
– MilkyWay90
Dec 19, 2018 at 3:02
\$\begingroup\$ @MilkyWay90 Nothing wrong with that, either. \$\endgroup\$
– Draco18s no longer trusts SE
Dec 19, 2018 at 3:23
\$\begingroup\$ Okay, thank you! I have another answer that you could crack (A BrainF one) \$\endgroup\$
– MilkyWay90
Dec 19, 2018 at 3:25
\$\begingroup\$ @MilkyWay90 I don't know anything about-- \$\endgroup\$
– Draco18s no longer trusts SE
Dec 19, 2018 at 3:28
\$\begingroup\$ Oh, okay. I didn't know anything about it until today so... \$\endgroup\$
– MilkyWay90
Dec 19, 2018 at 3:28

| Show 2 more comments

hyper-neutrino · Accepted Answer · 2021-05-20 20:43:08Z

1

Python 3, Makonede

class A:
    def __eq__(self, x):
        exit()

f(A())

Try it online!

The solution checks if x == x, and then loops while a or not a. This is always truthy. You cannot override the logical NOT operator so there is no way for this to be falsy. However, you can just input a custom object that exits the program when it is checked with == so the function is still called and runs but exits before it can hit the infinite loops.

answered May 20, 2021 at 20:43

hyper-neutrino♦

42.4k5 gold badges66 silver badges224 bronze badges

\$\begingroup\$ "there is no way for this to be falsy" your statement is falsy lol \$\endgroup\$
– Makonede
May 20, 2021 at 20:47
1

\$\begingroup\$ @Makonede oh. lol, that's smart \$\endgroup\$
– hyper-neutrino ♦
May 20, 2021 at 20:53
1

\$\begingroup\$ @Makonede an object which switches state when observed... who needs quantum computing when you have magic methods? \$\endgroup\$
– hakr14
May 20, 2021 at 21:05
\$\begingroup\$ @hakr14 i know. magical \$\endgroup\$
– Makonede
May 21, 2021 at 1:40

Add a comment |

emanresu A · Accepted Answer · 2021-05-22 00:58:55Z

1

Javascript, cracks Ethertyte's answer

f(eval('delete Object.prototype.hasOwnProperty'))

Try it online!

I can't believe I didn't think of this earlier.

answered May 22, 2021 at 0:58

emanresu A

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EasyasPi · Accepted Answer · 2021-05-22 03:26:02Z

1

Python 3, cracks hakr14's challenge

class x():
 def __init__(s):
  global isinstance
  isinstance=lambda x,y:0

f(x())

Try it online!

Simply replaces isinstance with a lambda returning 0, causing the isinstance call to always return 0, instantly exiting the loop.

edited May 22, 2021 at 3:26

answered May 22, 2021 at 1:52

EasyasPi

4,79015 silver badges22 bronze badges

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Stephen · Accepted Answer · 2017-07-24 18:07:58Z

0

JavaScript (ES6), programmer5000

"Infinity"

The condition first tests if the input has a .slice method (aka if it's a string or an array) and gets the absolute value of the numeric representation of the string/array. This is then added to 0 and checked against Infinity.

Try it online!

edited Jul 24, 2017 at 18:07

answered Jul 24, 2017 at 18:05

Stephen

13.8k3 gold badges52 silver badges116 bronze badges

\$\begingroup\$ :( I thought mine would last longer! Nice Job! \$\endgroup\$
– user58826
Jul 24, 2017 at 18:07
\$\begingroup\$ @programmer5000 thanks :) \$\endgroup\$
– Stephen
Jul 24, 2017 at 18:08
\$\begingroup\$ [Infinity] also works. \$\endgroup\$
– Cristian Lupascu
Jul 25, 2017 at 13:41
\$\begingroup\$ @w0lf also ["Infinity"] \$\endgroup\$
– Stephen
Jul 25, 2017 at 13:45

Add a comment |

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Will it halt? (Robbers)

Looking for uncracked submissions?

144 Answers 144

JavaScript, Step Hen

JavaScript, Step Hen

Python 3 (CPython), Veedrac

JS, Step Hen

R Jarko Dubbeldam

anyfix, Mayube

JavaScript by w0lf

R, Jarko Dubbeldam

CPython 3.6, by wizzwizz4

C++, SIGSEGV

C++, ecatmur

Python 3, Magenta

Python 3.5, Siphor

Swift, Alexander

Python 3: totallyhuman's post, 143 bytes

Python 2, Alex Hall

Python 3, Alex Hall

C (gcc), dj0wns

Python, Alex Hall

Python, by veedrac

Batch, SteveFest

Python 3, Alex Hall

JS, Step Hen

Python 3 by Veedrac

05AB1E, MilkyWay90

Python 3, Makonede

Javascript, cracks Ethertyte's answer

Python 3, cracks hakr14's challenge

JavaScript (ES6), programmer5000

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Python 3 (CPython), Veedrac

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anyfix, Mayube

R, Jarko Dubbeldam

C++, ecatmur

Python 3, Magenta

Swift, Alexander

Python 3: totallyhuman's post, 143 bytes

C (gcc), dj0wns

Python, Alex Hall

Batch, SteveFest

Python 3, Alex Hall

JS, Step Hen

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Python 3, Makonede

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JavaScript (ES6), programmer5000

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