Cipher Clock

[Elhuracan]

If you’re finding it hard to create a program that simulates the cipher wheel, try Googling “Wheatstone Clock Cipher” and you can find some stl files for free that you can edit 3D print to create a working spyclist cipher wheel.

[Louisfielding]

Well, other than building our own computer programme, it doesn’t really help as this cipher seems to be unique.

[Jamiecjx]

The clock cipher has a weakness in it that can be particularly exploited with the crib.

Bruteforcing the key is just a no because theres way too many possibilities.

A clue would to be to analyse what happens when you encrypt some random message but using different keys and see if a pattern is spotted.

[Elhuracan]

This blog has lots of helpful technical information if you want to make/computer simulate a Wheatstone device:
https://incoherency.co.uk/blog/stories/wheatstone-cryptograph.html

[18goyaan]

I am sssssssssssooooooooooo confused. I cant even utilise the piece of burnt paper

[18goyaan]

Dear Uncle Wilhelm.. DOESN’T WORK! PLS EXPLAIN! WHYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY!

[F6exb]

Because it is not a good crib!

[F6exb]

What a challenge!
I did a big part with just the clothes I’m standing up in (I don’t dare to use the French expression): paper, pencil, Word, Excel, and I see a little light. But sliding two pieces of paper with the alphabet and the encrypted key for hours is not very funny. If I want to get the plain text before 2021 challenge I need to learn a little more Python or PureBasic.

[Teymour_aldridge]

I’m not sure if it works (it doesn’t work for the clock cipher, but I think that’s a problem with my genetic algorithm) – I’m currently writing a hill-climbing algorithm for general clock ciphers.

My crossover function looks something like:

fn crossover(a: &str, b: &str) -> String {
    assert_eq!(a.len(), b.len());
    let crossover_points = rand::thread_rng().gen_range(0, a.len() / 2);
    let mut new: Vec<_> = a.chars().collect();
    for _ in 0..crossover_points {
        let number = rand::thread_rng().gen_range(0, b.len());
        new[number] = b.chars().nth(number).unwrap();
    }
    new.iter().collect()
}

It replaces a random number of positions in “a” with the corresponding values from “b”

[Jamieflet]

@Xantali I found the crib essential, it helped me to think of the what the operator is observing of the short and long hands as they encipher each letter. This helped me to to find a way to analyse the ciphertext and gather enough information to reduce the number of keys from 28! to 28. The rest fell into place very quickly then.

So… I have a solution and the key, but I am struggling to get my brain around what the keyword was. It certainly appears to me like there is one (should I be looking at 3B solution for this?), but has anyone actually managed to work back to this?

[Decoder-s]

Finally did it! A brilliant challenge. Don’t give up, it’s amazing how your brain will keep the problem ticking over while you do other things.

[Teymour_aldridge]

I’m writing a blog post on this, which I’m happy to share provided Harry approves. (it’s roughly 65% complete and walks through how to implement the algorithm)

There’s an interesting paper which gives some guidance.

https://eprint.iacr.org/2020/1492.pdf

Turns out climbing hills is better than genetic algorithms for this.

[F6exb]

I recover the key, thanks to the burnt paper. But as Madness and Kford-Academy says, it is also possible to recover it without a crib, by looking at bigrams and using a grid.

[Icybee]

@18goyaan Perhaps the message isn’t being sent to Uncle Wilhelm

[Chik2008]

What is this sorcery?!?

C = E(Ac, P)

P → E(Ac) → C

C = E(Ac, P) = M(Ac, S(P))

P→ S→M(AC)→C

(I got this from the latest case files and it makes no sense to me)

[Author]

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