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

You make a program and a regex.

For most inputs, one of the following happens:

  • the regex doesn't match, and the program returns a falsey value
  • the regex doesn't match, and the program returns a truthy value
  • the regex matches, and the program returns a truthy value

But for one (or more) input, the regex matches, but the program returns a falsy value.

You should try to keep this input hidden. Robbers will try to find an input where the regex matches, but the program returns a falsy value.

Post your program and your regex, as well as your language name, byte count of the program, regex flavor, and byte count of the regex.

If a robber finds an input where the regex matches, but the program returns a falsy value, even if it wasn't the intended input your program is considered cracked, and you edit in a link to the robber's crack.

If nobody has cracked your program & regex after a week, then edit in that your submission is safe and is therefore not eligible for cracking. Also edit in the intended input. The shortest (program bytes + regex bytes) non-cracked submission wins.


Let's say your regex is (all of the flavors):


And your program is (JS):


Then a valid crack would be 0, because the regex matches 0 but 0 is falsey.

Delimiters don't count in the byte count.

I can't enforce the banning of hash functions with rules, but please don't use SHA/RSA or anything that is unfair to the robbers

  • 2
    \$\begingroup\$ Are cryptographic hash functions allowed? I can imagine regex (..)+ and program print hash(input)==... \$\endgroup\$ – fireflame241 Aug 29 '17 at 1:21
  • 1
    \$\begingroup\$ @fireflame241 see edits \$\endgroup\$ – programmer5000 Aug 29 '17 at 12:39

Python, cracked

Regex: (?!.)

Program: from operator import not_

This is a little easier than my previous one.


JavaScript (Cracked)

Regex: ..
Program: a=>[...a].length-1

This is an easy one.

Intended solution



Python 3, cracked by ovs

Regex: .

Program: lambda a:len(a.lower()) == 1

This shouldn't be wasn't too hard.

  • \$\begingroup\$ Cracked \$\endgroup\$ – ovs Aug 29 '17 at 7:03

JavaScript ES6, 12 bytes + 3 bytes

Regex: ...

Code: a=>a+1<a-1+2

  • 2
    \$\begingroup\$ It is supposed to be hard to produce a false. With your program, almost any 3-digits number (and more inputs) would produce it. \$\endgroup\$ – Uriel Feb 18 '18 at 22:47

Java (OpenJDK 9), 198 bytes, (Cracked)

public static boolean isProgramValid(String n) {
    } catch(Exception e){
       return false;
    return true;

public static String regex = "^(?is)([a-z]+\\.){4}.*$";

Should be an easy one.

Try it online!

  • 3
    \$\begingroup\$ Isn't the question tagged as code-golf? Why all these white spaces and boolean instead of int? \$\endgroup\$ – user72349 Aug 29 '17 at 12:25
  • 1
    \$\begingroup\$ Cracked \$\endgroup\$ – PunPun1000 Aug 29 '17 at 12:33


Regex: .*

Program: !qC%.hz^TT"U|(¡

Try it online!

Simple hash function, shouldn't be allowed, but it is. I could've made it much stronger, but I tried to optimize for byte count.

  • 1
    \$\begingroup\$ Stop suggesting we make things into pop-cons! Pop-cons are almost never good nowadays, and this challenge is much much better as a cops and robbers. The issue is that the OP hasn't yet clarified, which does not mean that it should be a pop-con. \$\endgroup\$ – caird coinheringaahing Aug 29 '17 at 11:50
  • \$\begingroup\$ He has to do one of the two things, otherwise the winner is whoever can create the shortest hash that remains secure. \$\endgroup\$ – Sebastian Aug 29 '17 at 11:58
  • 1
    \$\begingroup\$ I'm not arguing with that, I'm arguing with the Maybe make the challenge a popularity contest instead. That is in no way the correct solution to a problem in a question. Take a look through this and see how many are closed. Pop-con is not a solution to an under specified challenge, it is a separate challenge type. \$\endgroup\$ – caird coinheringaahing Aug 29 '17 at 12:07
  • \$\begingroup\$ Good point, turns out I wasn't aware of what makes a challenge a good popularity contest. Edited my answer. \$\endgroup\$ – Sebastian Aug 29 '17 at 15:23

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