This seems to be a good solve. It’s about the constellations.

pathantheeclipticpa
ssesthroughthirteen
consteelationsthetw
elvetraditionolgadi
acconsteelationsplu
sophimchuswhichinte
rfectsbetweenscorpi
aandsagittoriustwel
vesignsofgodiacaref
irstariessecondtamr
usthirdgemininourth
concerfiftheeasifth
virgoseventhlibraei
ghthscorpioaddition
alaphimchusninthsag
ittoriustenthcapric
orneleventhadmarius
tweenthpiscesgodiac

Jarlve,
That’s pretty close, congrats! The second part is still mostly unreadable, but it uses all of the actual names of zodiac signs, so that’s understandable. What did it take? I’m guessing you used AZdecrypt, right? How many iterations? I tried 10,000,000,000 (i.e. 10 Billion) and didn’t get anything even close to your solve.

By the way, if I can make a suggestion for AZdecrypt. From what I gather, you use "THE ADVENTURES OF SHERLOCK HOLMES" to build 4-gram stats and then score the plaintext based on that, correct? That probably explains why your solve above was still so far from actual plaintext — since your corpus likely didn’t use any of the zodiac sign names. You can improve the solve rate quite a bit by using a more comprehensive 4-gram stats from a bigger corpus, and you can get an even bigger improvement if you switch to 5-gram stats. You can get that data from here, for example:
http://practicalcryptography.com/crypta … equencies/

hey daikon,

I used a version of AZdecrypt that is in development which can use up to 6-grams drawn from a new 90 megabyte corpus (Project Gutenberg books). 5-grams were used to solve your second cipher. Yes, it would probably be worthwhile to draw from even a bigger and more diverse source. 4-grams are used for all current versions of AZdecrypt out there because they provide a very good ratio between solving power and speed. I like the website Practical Cryptography allot and have used its information since the start of writing the first iteration of my solver.

I don’t think the second part is mostly unreadable.

pathan the ecliptic passes through thirteen consteelations
the twelve traditionol gadiac consteelations plus ophimchus
which interfects between scorpia and sagittorius twelve signs
of godiac are first aries second tamrus third gemini nourth
concer fifth eea sifth virgo seventh libra eighth scorpio addition
alaphimchus ninth sagittorius tenth capricorn eleventh admarius
tweenth pisces godiac

Jarlve,
Fair enough, apparently I’m not too good at parsing continuous streams of letters into English words. Probably just not enough experience doing that over and over again, compared to you. 🙂

Using 6-grams to score solves would be awesome and I think should improve AZD even more. Although I think 90 Mb corpus would be too small. Practical Cryptography data used 4+ Gb corpus. Although for some reason if you add up all counts for 3-grams you get 4,274,127,909, for 4-grams you get 4,224,127,912, and 5-grams = 4,174,127,916. Perhaps they didn’t cross sentence boundaries when counting N-grams, so you get higher counts for lower Ns?

By the way, which algorithm are you using in AZD? Hill-climb? Simulated annealing? Genetic algorithm? Something else?

daikon,

This is a new solve using 5-grams from Practical Cryptography.

path of the ecliptic passes through thirteen constellations
the twelve traditional bodiac constellations plus ophimchus
which interfects between scorpio and sagittarius twelve signs
of bodiac are first aries second tamrus third gemini fourth
cancer fifth leo sinth virgo seventh libra eighth scorpio addition
alophimchus ninth sagittarius tenth capricorn eleventh atmarius
twelfth pisces bodiac

It’s certainly better and on first impression cipher recovery rate improved as well. Your tip has panned out, thank you. Though the program loses the flexibility to manipulate n-grams at start-up, for instance slightly randomizing or removing characters from the corpus. Which has proven valuable for smokie_treats’s wildcard hypothesis.

I don’t want to go to much into program details but it’s very similar to simulated annealing (performs about the same also) and it was something I came up with myself before I even learned of SA. In general my program has not much intelligence and relies on it’s speed.

I’m glad to be of some small help with my suggestions. 🙂 Using 6-grams should improve the solves even further, and 7-grams are not that impractical. If you keep each 7-gram score to 1 byte which should be plenty for a log score, you "only" need 8,031,810,176 bytes (26 to the power of 7) to keep the 7-gram stats in memory in an optimized for speed array, or less than 8Gb of RAM total. Not entirely out of the realm of possible for modern computers. How much improvement you get from 7-grams vs 6-grams is still remains to be seen, as it will greatly depend on the size of your corpus at that point.

I don’t want to go to much into program details but it’s very similar to simulated annealing (performs about the same also) and it was something I came up with myself before I even learned of SA. In general my program has not much intelligence and relies on it’s speed.

You don’t actually need "much intelligence" to solve many problems. If you think about it, hill-climb algorithm is very dumb to begin with. I always thought that in the simplest terms it sounds just like how a 5-year old would approach solving a problem. 🙂 You just nudge the solution a bit in a random direction and if it gets better, you keep it, otherwise you go back to the old solution and nudge some more. What could be simpler or dumber than that? And yet, it can solve a wide range of very complex optimization problems. It just takes a lot of nudging and a few clever improvements to speed up the whole process. So not much intelligence and a lot of speed is a *very* good thing.

Ok, here’s my third, and hopefully last, attempt at unsolvable ciphers. Even though the previous cipher already doesn’t get solved by any of the currently available auto-solvers, the next version of AZD that Jarlve is working on was able to crack it. Let’s see if this cipher proves to be unsolvable:

vKeO<]r4]GdgRq<aa@
tHZ_LWAo1MiXBI[2r3
+gCYNSDnE^T56P+JUF
_K+7Xf+bV8cGm9+ZoA
kde@a1q2+gBl3+CnQ5
r6+Rm7[Dp8+O]4<EYL
WFo9Mi^GIhNq@gO+hH
SZnJ+1P+KTA_d+2Xf+
bU3cBk5+CoDlHe64iY
EgF^LVGn7Mi_ZQm8RA
a<s9p@WSo1[P<BXNTC
n2OiYDIE^4UFoG_V35
ghJWZn6LiXAQ+KSBYd
+7^f+bT8cCk9+DoElH
eF_MUG@1+ghNhZnhJf
VRm2[AP<+hW<+]OQ]K
BX4SCoR3r5+IhdTDn[
6q7+g+L+8E9@+P+HUF
YJ+1^f+bV2cGk3+ZoA
lKeBm5+CnQ6r7+gDk8
+EohMq9Rl@[Z0dI4<]

I even managed to mimic Z340 somewhat, with an abundance of pluses and a lovely signature at the end, which is *not* a random filler. In theory it should be much easier to solve, as it is quite a bit longer than Z340, and nearly the length of Z408, if you trim the random filler at the end of Z408. Just as before, this is a straight homophonic substitution cipher. Every ciphertext symbol translates to a single unique plaintext letter. The message is in plain English, with the number of rare words kept to an absolute minimum – 5, each of which is a proper noun, and I allowed myself only one spelling mistake, to be consistent with Zodiac’s writing style. The content of the message is entirely my own, but it would not be unexpected for Zodiac to write something similar. Just as before, I’ve also confirmed that the plaintext does get solved by ZKD and AZD if I lower the number of unique symbols in the ciphertext (i.e. if the multiplicity is improved).

So it seems I was able to stump all current auto-solvers after all? And only on the third try. I’m crackproof! 🙂 I’ll wait a couple of days before revealing the plaintext, just in case someone is still trying.

Yes, I can’t come up with a solve. Though your 3rd cipher scores much higher than the 340. I don’t know what you did but well done!

Edit: I think you tried to induce high multiplicity by assigning some high counts to a few symbols and low counts to the bulk. It’s something which I have thought of aswell in relation of the 340 and that we may need to come up with a better calculation of the cipher difficulty then multiplicity.

I’ve worked on this a bit as well, and have no solve. But it’s a weird cipher. Daikon, in your first post here, you said:

And I’m not talking about carefully crafted plaintext that doesn’t have certain common letters (like passages from the novel "Gadsby"), or that has an abundance of rare letters, or a very high, or very low IoC.

So, in this one, the + symbol must be either E, T, A, O, I, or N. Assuming you didn’t change the frequencies too much. And based on the patterns that ZKD finds, there are a fair number of repeats. Very odd.

You got me, though. Good job. You might as well post the plaintext, I think me and Jarlve are the only people here that work on these things.

-glurk

EDIT: Just to add, this is honestly the strangest cipher I’ve ever seen. If it turns out to be fairly normal English text, I bow down to you.

Maybe the repeats/patterns can be explained by the plaintext being very repetitive. The same thing repeating over and over again, or use of a very limited vocabulary.

Edit: that would also artificially raise the multiplicity of the cipher, I believe. Yes, I think this is what daikon’s guilty of!

Here’s a row with a spike in columnar repetitions:

I did some experiments and I believe Jarlve is correct. There is a lot of repetition in the plain text.

Daikon, don’t post the solution yet, I think we all need the challenge to be extended a little.

OK, I cracked it (daikon, please confirm the solution I pm’d you), but it required many manual steps. Hint: Reduce the multiplicity before trying to crack this.

It will be interesting to determine how to get an auto-solver to uncover this kind of plaintext.