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

Kpprvugdnoavwugzjdvwqswwcvoburnxvlqqrktduhvkkvkvobhltvvwkpgdpbydalylnovuawqoqrmiquusgonlpjgutrtvdxvlcpprvsguhhewcqflowaskqirpdrkqqgdpgvkkvkvconrphuwtlpjurcxvrertugfvluxuhnhuvkjwhuvkfqxngjdxhyukwvhplvrwwprtpconbcqfwjhpjqwvhpukgqivkgvrdehuzjbflfqvlfrvkcwvkkvkvihvwkqisthvwaoqqivqlylnogqflvkgug

Whyvhfpnrfnejnffgnoorqgjraglguerrgvzrfnppbeqvatgbebznauvfgbevnanaqculfvpvnafhrgbavhf

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

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    $\begingroup$ 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
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    $\begingroup$ 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
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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

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  • $\begingroup$ 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

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