# Code Wheel

## by IWorkInPixels, published

## Description

My dad taught me this cipher when I was a kid, and I've wanted for a long time to make a printable code wheel that would allow others to learn the cipher. This is some sort of variant on the Vigenere cipher (http://en.wikipedia.org/wiki/Vigenere_cipher) but I'm not sure of the actual name of this variant, so if anybody knows, please let me know in the comments! I know it is also related to the key autokey and Alberti ciphers.

In addition, I have a prize for the first person to correctly decrypt the ciphertext below and post the plaintext to the comments... I loved the steganography challenge in the Seej block, and it's high time we had another contest... Good luck!

Oh, and you will find that brute force is not the way to go... there are 1.243841e 142 possible keys. :)

UPDATE: There is now a monocase alphas only ciphertext of the same plaintext, which reduces your keyspace down to 4.0329146e 26 possible keys.

Keep in mind that an attacker doesn't need to do an exhaustive search of all of these keys; once he has some of the letters right, they will help him get more letters right, and the effect will snowball until he has the entire key. Indeed, I cracked the alphas only version by hand in less than 2 hours.

**This algorithm should therefore not be used for anything that you actually need to keep secret.**

[BEGIN CIPHERTEXT]

-G},x3Pp__DPpqSk6'iF5]E0ww"QhV.yEei{b>~wu2)5ecNGGO^AF3w.ur!RE ]E>}.

[END CIPHERTEXT]

[ALPHAS ONLY CIPHERTEXT]

TKAXPDDFJCJFJIOCXJIIOENGQACFOVDBDFULLYWCFAVYKFOLEWP

[END ALPHAS ONLY CIPHERTEXT]

I design things in my free time, just for fun. If you like my stuff, send me some DOGE. It'll totally make my day! Thank you.

D9hPusbPPdbUXgmsRqRcb7jRisVLnriCuP

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## Instructions

1: Print it out.

2: Put it together.

3: Follow the instructions on the webpage download. If you want, you can also use this wheel to perform simple Caesar ciphers.

4: Optional: Use the .SCAD file to print two copies of a new key, and give one to a friend. Now you can communicate securely with that friend!

5: If you'd like to participate in the contest, you will want to download the python version of the cipher.

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wheel matches a given letter on the outer wheel, then the letter on the

inner wheel which lines up with the A on the outer wheel + the letter on

the inner wheel = the key length.

In the photos above, for example, M and O are shown to each line up with the A on the other wheel.

In the case of the printed wheel, the key length is 26. So M + O = 26.

Some

of these combinations are more likely to appear in english text than

others, but there are only 52 of them to check, so worst case scenario, a

script could pretty easily brute force its way through all of them, and

either look for English words itself or show the results to you to pick

a winner.

Lather, rinse, repeat until you have enough of the key to solve the puzzle.

If you see the word "MOM" in the ciphertext, it means one letter was encoded as an M, the next letter moved x distance around the circle and landed on the O, and the third letter moved 26-x distance around the circle and landed back on the M again. The first letter could be anything, but there are only a small number of possibilities for what the 2nd and 3rd letter could be, and it is therefore easy to check your guesses.

abrahamlincolnonline.org/lincoln/speeches/gettysburg.htm

with some filler at the beginning and end so I'm not just giving away the key. You now have a long piece of ciphertext encrypted with the same key (though not the same rotation of the key) as the original contest message, and you now have the mother of all cribs. Enjoy!

So I haven't had time to really develop the two cypher rings that you credited as parent object of this (thanks for that) but generally letters don't print well laying down like that. I'm not sure the scale but if they're not at least 1mm thick all around there's not much chance of this printing successfully. I've tried getting around this by printing letter rings on sticker paper and sticking it to the rings, but that doesn't always work because stickers don't like plastic unless you corona treat them, which not everyone has a plasma generator to do that with. So I've kind of back-burnered that project for a little while.

As for the ciphertext above, it was indeed written by this cipher, although I have added many more characters to the alphabet than the ones that will fit in the 3 inch printed version. I have already given away the key space (which is alphabet size factorial), and therefore also the alphabet size and key length. I will begin posting more hints this afternoon. I will also upload the python code used to create my ciphertext.

It *is* possible to crack the cipher, but it will require some research into some cryptanalysis methods that were declassified by the NSA in 2005, and are available on their website.

Good luck!

If you *really* want to try cracking it using only pen and paper, here's the same message, encoded using the printable code wheel, alphas only (spacing and punctuation removed to hide plaintext word lengths):

TKAXPDDFJCJFJIOCXJIIOENGQACFOVDBDFU LLYWCFAVYKFOLEWP

If you see the word "MOM" in the ciphertext, it means one letter was encoded as an M, the next letter moved x distance around the circle and landed on the O, and the third letter moved 26-x distance around the circle and landed back on the M again. The first letter could be anything, but there are only a small number of possibilities for what the 2nd and 3rd letter could be, and it is therefore easy to check your guesses.