This code requires one half to solve the other. This shouldn't be too difficult.



Yeah, this shouldn't take anybody more than ten minutes now that I look back on it. Enjoy!

  • 1
    $\begingroup$ Welcome to our site! As is, this question sadly not a very good fit, but don't let that discourage you. Please take a glance over this informative posting and see if you can edit your puzzle to be better received. $\endgroup$ – Sconibulus Dec 30 '16 at 14:55
  • 1
    $\begingroup$ Thank you! This is my first time here, so I've been looking for something like that. The revised version of this will be here soon. $\endgroup$ – Jun Hayakawa Dec 30 '16 at 15:00

The first half is:

VIGENERE CIPHER (key:CD) or GRONSFELD CIPHER (key:23) I'm not really sure what to put as my solution phrase. This is my first time anyway. I will try to look for spelling errors but I am not perfect and I'm typing on a phone and this is all one string so autocorrect is useless. I guess I could have written it out normally and then gotten rid of the spaces. Why didn't I do that? This is getting pretty long so I will end it here.

The second half is:

CAESER CIPHER (ROT13) Julius Caesar was stabbed twenty-three times according to Roman historian and physician Suetonius.

A Vigenere cipher is a multi-dimensional Caesar cipher. The offset value follows a key which repeats. To encode on a CD key, you alternate shifting each letter +2 and +3. To decode, it's the reverse: you alternate shifting each letter -2 and -3.

I have modified this Vigenere square as a decoding square:

vigenere decoding square

  • $\begingroup$ How have you found out the Vigenère key, CD or CDCD? $\endgroup$ – M Oehm Dec 30 '16 at 15:50
  • $\begingroup$ Well, almost correct. I had used the Gronsfeld cipher, and 23 had been the key. $\endgroup$ – Jun Hayakawa Dec 30 '16 at 15:51
  • $\begingroup$ I added more information about how Vigenere ciphers work, and how they are decoded. My Python code always checks through all of the 2-letter keys first, so CD actually came up pretty quickly. I compare letter distribution to actual English distribution and if its close, the decoded string prints out. Brute force at its best (or worst)... $\endgroup$ – wildBillMunson Dec 30 '16 at 16:39
  • $\begingroup$ I edited my answer to include Gronsfeld 23, bringing it in line with the OP's specific method of encryption. $\endgroup$ – wildBillMunson Dec 30 '16 at 16:47
  • $\begingroup$ Any reason my answer hasn't yet been accepted? Am I missing something? $\endgroup$ – wildBillMunson Dec 31 '16 at 21:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.