Hello, today I have decided to stop working on z340.

O.k. Hopefully you will be back. I was thinking about writing how much I appreciated your work. You really need a partner, someone to work with.

I get burned out too. It is like gambling, but instead of with money with time and mental energy. The odds are small, but if we win we will be a hero, and get to find out what he said, and have closure. If we lose it is painful until we get over it, start thinking about it more and more, then get an idea and endeavor to try the idea. Then fail, get upset, burned out, and swear to never try again. Then… slowly but surely… start thinking about it again.

Pivots are quite common in plaintext (I posted some numbers above). If plaintexts of length 340 are shuffled randomly, the probability for pivots remains unchanged. The more cyclically a homophonic substitution is applied to these plaintexts, the more often the pivots disappear.

That is interesting. Probably because the symbols that are closest to each other are not the same because of the cycles. The symbols that are farther apart are probably more often the same with more cycles. It has to do with the plaintext frequency and the key, but I can totally see how that and randomness would affect the pivots. Makes sense and thanks for doing that. Every little bit that we understand is a step forward. Thanks for the contribution.

I stick with the fact that the pivots were probably inserted intentionally, in order to lure by patterns on a wrong track. After the patterns were created, the remaining plaintext was substituted. For me this is a plausible explanation for the incomplete cycles in z340. Especially around the pivots there are conspicuously many patterns. The trigrams IOF and FBC (inverted C) are also directly involved.

I used to think that the pivots were a red herring also, but have recently changed my mind because of P29. What are the odds of P29 being created by the cipher, but him putting the pivots also on P29? Seems rather improbable, and the other symbols in and around the pivots make partial smaller pivots. But maybe you are right.

EDIT: Let’s make a simple rule. The pivots have to be whole, they cannot be broken by the message borders. So that means 14 columns x 17 rows, or 238 possible positions for the first pivot. Then, 237 possible positions for the second pivot. So, odds are 1 in 237 that he would have put the pivots on P29, assuming he did not know about P29?

But before I start thinking again, I better stop.

Good Luck!

today I have decided to stop working on z340. I had a great time here in the forum and would like to thank everyone. It is nice to exchange ideas and knowledge and work together on a solution. I was able to learn a lot and don’t see any minute spent with z340 as a waste of time. Anyone who starts such a hobby knows that it maybe will not bear fruit. Nevertheless, I was able to profit enormously from this time. I always enjoyed the atmosphere here in the forum, because there was always a respectful and friendly tone. Once again, thank you for that.

Thank you for everything Largo. Know that you are always welcome.

Oh! I figured it out! In the "code box" it scrolls horizontally! Thats odd. But hey, as long as it’s there.

You can hit return/enter after each line of the cipher, to make it format better. Like this:

VYMIRGITRCDUZIEOUFLT
AELSPHNDFRNVBQIIRGIO
MTPNCBUAITIMAROUSTAU
TJHSKDGNRDBABOZLNRUP
ENMAESLXTNRHGBOIAELH
MCYULEIECPTOSRMBCPEA
NQLZAESMRDAZAESGCLEK
SYGHNSNETOIAPJDMSCRS
ARFERFSNFSIOQVCIATUF
CELOESSPUVZITQNNOXT

Awesome.

Again, thank you for your help and patience, I love ciphers and codes, but I am not very adept in technological matters, even in my chemistry work I still prefer an old fashion notebook and pen to a computer device.

I am an autodidact in most every subject, so pardon any seemingly neophyte errors, I am a quick learner and appreciate when my mistakes are corrected. I do know a good deal, but there is also a good deal I have yet to learn, as I said I am a very quick learner, but I am also my only teacher…

Any way, thank you for all the help, and thank you for your patience in helping me.

but I am also my only teacher…

Join the club!

A club of autodidact cipher enthusiasts.

Looks like my monoalphabetic decryption is missing a character…

I’ll have to review and correct it.

Been making slow progress otherwise.

monoalphabetic decryption

How did you arrive at it?