The greedy regex (cops)

Question

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.

Example

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

/^[0-9]$/

And your program is (JS):

x=>Boolean(parseInt(x))

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

• fireflame241: Are cryptographic hash functions allowed? I can imagine regex (..)+ and program print hash(input)==...

Answer 1

JavaScript (Cracked)

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

This is an easy one.

---

Intended solution

😊

• Eli Richardson: Cracked

Answer 2

Python, cracked

Regex: (?!.)

Program: from operator import not_

This is a little easier than my previous one.

Answer 3

JavaScript ES6, 12 bytes + 3 bytes

Regex: ...

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

• Uriel: It is supposed to be hard to produce a false. With your program, almost any 3-digits number (and more inputs) would produce it.

Answer 4

Python 3, cracked by ovs

Regex: .

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

This (削除) shouldn't be (削除ここまで) wasn't too hard.

• ovs: Cracked

Answer 5

Pyth

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.

• caird coinheringaahing: 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.
• Sebastian Mendez: He has to do one of the two things, otherwise the winner is whoever can create the shortest hash that remains secure.
• caird coinheringaahing: 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.
• Sebastian Mendez: Good point, turns out I wasn't aware of what makes a challenge a good popularity contest. Edited my answer.

Answer 6

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

public static boolean isProgramValid(String n) {
 try{
 Class.forName(n);
 } catch(Exception e){
 return false;
 }
 return true;
}
public static String regex = "^(?is)([a-z]+\\.){4}.*$";

Should be an easy one.

Try it online!

• user72349: Isn't the question tagged as code-golf? Why all these white spaces and boolean instead of int?
• PunPun1000: Cracked